## Solid Geometry Chapter 2 Theorems Related To Angles In A Circle

## Angles In A Circle

Let ABCD is a. circle and O is its centre. Points A, B, C and D are on the circle.

If A, D and B, D are joined, then there is produced an angle ∠ADB. Then ∠ADB is called the angle in the circle produced by the arc \(\overparen{\mathrm{AB}}\).

Since D lies on the circumference, the angle ∠ADB is also called angle in circumference.

Similarly, ∠BAC, ∠ACB, ∠BDC, ∠CAD are all angles in a circle, which are produced by the arcs \(\overparen{\mathrm{BC}}, \overparen{\mathrm{AB}}, \overparen{\mathrm{BC}}, \overparen{\mathrm{CD}}\) respectively.

**WBBSE Solutions for Class 10 Maths**

Hence, any angle on the circle or on the circumference of a circle which is produced by any arc of the circle is called the angle in a circle.

Observe that by joining the two end points of an arc with a point situated on the opposite side of the arc and also on the circumference of the circle, we can get an angle in a circle.

However, since there exists infinitely many points on the opposite side of the fixed arc, we can draw an infinite number of angles’ in a circle.

**Wbbse Class 10 Maths Solutions**

Thus the number of angles in a circle produced by a fixed arc is infinite.

Again, the chord produced by joining the two end points of an arc, we can construct more than one angle in a circle on both the sides of this chord.

Such as, in the given two angles in a circle are ∠ACB (on the upper side) and ADB (on the lower side) produced by the arc \(A \overparen{D D} \text { and } \overparen{A C B}\) respectively.

** Characteristics of angles in circle **

- The vertex of angle in circle always lies on the circumference of the circle.
- The centre of the circle may or may not within the region of the angle in circle.
- Any angle in circle must be produced by an arc of the circle.
- Infinite number of angles in circle can be made by an arc of the circle.
- Angle in circle may be either acute, obtuse or right angle.

Let ABC be a circle with centre at O. \(\overparen{\mathrm{AB}}\) is any arc of the circle.

Now, joining A, O and B, O by straight lines we get an angle ∠AOB at O. This angle ∠AOB is called the central angle.

Hence the front angle produced by an arc of the circle is called the central angle.

In the given ∠AOB, ∠BOC, ∠BOD are all central angles, which are produced by the arcs \(\overparen{\mathrm{AB}}, \overparen{\mathrm{BC}} \text { and } \overparen{\mathrm{BD}}\) respectively.

It must be noticed that both ∠AOB and reflex ∠AOB are produced by the arc AB.

**Wbbse Class 10 Maths Solutions**

Hence two and only two central angles can be constructed by a certain arc one of which is a reflex angle.

** Characteristics of central angles **

- The vertex of any central angle always lie on the centre of the circle.
- Two and only two central angles can.be constructed by a fixed arc of a circle, one of which is a reflex angle.
- The value of a central angle may be either an acute or an obtuse or a right-angle or a reflex angle.
- The range of a central angle is from 0° to 360° including the both.
- The central angle produced by the entire circumference of a circle is 360° and that produced by the circumference of a semi-circle is 180°.

**Wbbse Class 10 Maths Solutions**

** Relation between angle in circle and its central angle produced by any arc of a circle **

In the above discussion, we have seen that by any arc of a circle both angle in circle and central angle can be constructed, So, it is the question that is their any relation between them?

Let the central angle produced by an arc AB of a circle with centre at O be ∠AOB and ∠ACB be its angle in circle.

If we measure these two angles ∠AOB and ∠ACB by a protractor, we shall see that ∠AOB = 2 ∠ACB, i.e., the angle which an arc of a circle subtends at the centre is double the angle which it subtends at any point on the remaining part of the circumference.

We shall now logically prove this theorem by geometric method.

## Solid Geometry Chapter 2 Theorems Related To Angles In A Circle

## Angles In A Circle Theorem

**Theorem: Prove that the front angle formed at the centre of a circle by an arc, is the double of the angle formed by the same arc at any point on the circle.**

**Given:** ∠AOB is the central angle of the circle with centre at O, produced by the arc APB and ∠ACB is the angle at any point on the circle formed by the same arc APB.

**Wbbse Class 10 Maths Solutions**

**To prove:** We have to prove that ∠AOB = 2 ∠ACB.

According to the length of the arc APB, these may be of three types, In Number (1) and (4), APB is a minor arc, In number, (2) APB is a semi circle and in number (3), APB is a major arc.

We shall have to prove that theorem for all the cases.

**Construction:** Let us join C. O and CO is produced upto D.

Proof: In ΔAOC. OA = OC (radii of same circle)

∴ ∠OCA = ∠OAC [in all the mages]

Again, tor the ΔAOC, in every cases, external ∠AOD = internally opposite (∠OAC + ∠OCA)

= ∠OCA + ∠OCA [∠OAC = ∠OCA] = 2 ∠OCA …….. (1)

Again, in ΔBQC, OB = OC [radii of same circle]

Similar is the case, when CO of ΔBOC is produced upto D, external ∠BOD = internally opposite (∠OCB + ∠OBC)

= ∠OCB + ∠OCB [∠OBC = ∠OCB]

= 2 ∠OCB……….(2)

Now in the cases of (1) arid (2), ∠AOD + ∠BOD = 2 ∠OCA + 2 ∠OCB

or, ∠AOB = 2 (∠OCA + ∠OCB) or, ∠AOB = 2 ∠ACB

In the (3), ∠AOD + ∠BOD – 2 ∠OCA + 2 ∠OCB

or, reflex ∠AOB = 2 (∠OCA + ∠OCB)

or, reflex ∠AOB = 2 ∠ACB

In the (4), let us subtract (1) from (2) to get,

∠BOD + ∠AOD = 2 ∠OCB + 2 ∠OCA

or, ∠AOB = 2 (∠OCB – ∠OCA) or, ∠AOB – 2 ∠ACB

∴ in every cases, ∠AOB = 2 ∠ACB (proved)

## Solid Geometry Chapter 2 Theorems Related To Angles In A Circle

## Angles In A Circle Multiple Choice Questions

**Example 1. In the given, O is the centre of the circle and PQ is one of its diameter, then the value of x is **

- 140
- 40
- 80
- 20

Solution: In the given, ∠POR = 140° (Given),

∴ ∠ROQ = 180° – 140° = 40°

Now, the angle in circle produced by the arc \(\overparen{Q R}\)

= ∠QSR and the central angle = ∠ROQ.

∴ ∠QSR = \(\frac{1}{2}\)∠ROQ [by theorem] = \(\frac{1}{2}\) x 40° = 20°

∴ x° = ∠QSR = 20°, ∴ x = 20,

∴ 4. 20 is correct.

**The value of x is 4. 20.**

**Example 2. In the given, if O is the centre of the circle, then the value of x is**

- 70
- 60
- 40
- 200

Solution: In the given, ∠POQ = 140°, ∠POR = 80°

∴ ∠QOR = 360° – (∠POQ + ∠POR)

= 360° – (140° + 80°) = 360° – 220° = 140°

Again, the angle in circle produced by the arc \(\overparen{Q R}\) = ∠QPR = x° and the central angle = ∠QOR = 140°

∴ ∠QPR = \(\frac{1}{2}\)∠QOR [by theorem]

= \(\frac{1}{2}\) x 140° =70°, x° = 70° ∴ x = 70

∴ 1. 70 is correct.

**The value of x is 1. 70.**

**Example 3. In the given, if O is the centre of the circle and BC be any diameter of it, then the value of x is**

- 60
- 50
- 100
- 80

Solution: In the given, ∠OAB = 50°, ∠ADC = x°,

Now, OA = OB (radii of same circle)

∴ ∠OAB = ∠OBA – 50° [∠OAB – 50°]

∴ ∠AOC = internally opposite (∠OAB + ∠OBA) = 50° + 50° = 100°

Again, the angle in circle produced by the circular arc \(\overparen{A C}\) = ∠ADC = x° and central angle = ∠AOC = 100°

∴ ∠ADC = \(\frac{1}{2}\) ∠AOC [by theorem]

or, ∠ADC = \(\frac{1}{2}\) x 100° [∠AOC = 100°] .

. or, x° = 50°, x = 50,

∴ 2. 50 is correct.

**The value of x is 2. 50.**

**Example 4. O is the circumcentre of triangle ABC. If ∠OAB = 50°, then the value of ∠ACB is**

- 50°
- 100°
- 40°
- 80°

Solution: Let us join O, B.

Now, OA = OB (radii of same circle)

∴ ∠OBA – ∠OAB = 50° (Given)

∴ ∠AOB = 180° – (∠OBA + ∠OAB) = 180° – (50° + 50°) = 180° – 100° = 80°

Now, the angle in circle produced by the arc \(\overparen{Q R}\) = ∠ACB and the central angle = ∠AOB = 80°

∴ ∠ACB = \(\frac{1}{2}\) ∠AOB [by theorem] = \(\frac{1}{2}\) x 80° = 40°

∴ 3. 40° is correct.

**The value of x is 3. 40° .**

**Example 5. In the given, if O is the centre of the circle, then the value of ∠POR is **

- 20°
- 40°
- 60°
- 80°

Solution: In the given, ∠OPQ = 10°, ∠ORQ = 40°

∴ OP = OQ [radii of same circle]

∴ ∠OQP = ∠OPQ = 10° (Given)

Again, OQ = OR (radii of same circle)

∴ ∠OQR = ∠ORQ = 40° (Given)

Now, ∠PQR = ∠OQR – ∠OQP [according to the image]

= 40° – 10° = 30° [∠OQR = 40° and ∠OQP = 10°]

Then, the angle in circle produced by the arc \(\overparen{P R}\) = ∠PQR = 30° and the central angle = ∠POR

∴ ∠POR = 2 ∠PQR [by theorem]

. = 2 x 30° [∠PQR = 30°] .

= 60°

∴ 3. 60° is correct.

**The value of x is 3. 60°.**

## Solid Geometry Chapter 2 Theorems Related To Angles In A Circle

## Angles In A Circle True Or False

**Example 1. The number of angles in a circle produced by an fixed arc is infinite.**

Solution: True

**Example 2. The vertex of angle in circle does not always lie on the circumference of the circle.**

Solution: False.

## Solid Geometry Chapter 2 Theorems Related To Angles In A Circle

## Angles In A Circle Fill In The Blanks

**Example 1. On the same circular arc, the angle in circle is _______ of its central angle.**

Solution: Half [by theorem]

**Example 2. The lengths of the chords AB and AC of the circle with centre at O are equal. If ∠APB and ∠AQC are two of its angle in circle, the values of the angles are ______**

Solution: Equal

**Example 3. If O be the circum-centre of an equilateral triangle, then the value of the front central angle produced by any one of its sides is equal to ______.**

Solution: 120°, since, let ABC be an equilateral triangle.

∴ ∠BAC = ∠ABC = ∠ACB = 60°.

Now, the front central angle produced by BC = ∠BOC.

The angle in circle produced by BC = ∠BAC.

∴ by theorem, ∠BOC = 2 ∠BAC = 2 x 60° = 120° [∠BAC = 60°]

## Solid Geometry Chapter 2 Theorems Related To Angles In A Circle

## Angles In A Circle Short Answer Type Questions

**Example 1. In the adjacent O is the centre of the circle. If ∠OAB = 30°, ∠ABC = 120°, ∠BCO = y° and ∠COA = x°, then find the values of x and y.**

Solution:

**Given :**

In the adjacent O is the centre of the circle. If ∠OAB = 30°, ∠ABC = 120°, ∠BCO = y° and ∠COA = x°

According to the given, ∠AOC = x°, ∠ABC = 120°, ∠OCB = y°, ∠BAC = 30°

Now, the central angle produced by the arc AC = reflex ∠AOC and angle in circle = ∠ABC =120°

∴ reflex ∠AOC = 2 x ∠ABC [by theorem] = 2 x 120° – 240°

Then ∠COA – 360° – reflex ∠AOC = 360° – 240° – 120°

∴ x° = 120° ⇒ x= 120

Again, ∠AOC + ∠OCB + ∠OAB + ∠ABC = 360 [the sum of 4 angles of a quadrilateral is 360°]

or, 120° + y° + 30° + 120° = 360°

or, y° + 270° = 360° or, y° = 360° – 270° or y° = 90°

∴ x = 120 and y = 90.

** The values of x and y 120 and 90.**

**Example 2. O is the circumcentre of AABC and D is the mid-point of BC. If ∠BAC = 40°, then find the ****value of ∠BOD.**

Solution:

**Given :**

O is the circumcentre of AABC and D is the mid-point of BC. If ∠BAC = 40°

D is the mid-point of BC, BD = CD

Now, in triangles ΔBOD and ΔCOD, OB = OC [radii of tire same circle], BD = CD and ∠OBD = ∠OCD [OB = OC]

∴ Δ BOD = Δ COD [by the condition of S-A-S congruent]

∴ ∠BOD = ∠COD [similar angles of congruent triangles]

∴ ∠BOC = 2 ∠BAC [by theorem]

= 2 x 40° [∠BAC – 40°] = 80°

Then from(1) we get, ∠BOD = \(\frac{1}{2}\)∠BOC = \(\frac{1}{2}\) x 80° [∠BOC = 80°] = 40°

**The ****value of ∠BOD = 40°**

**Example 3. Three points A, B and C lie on a circle with centre at O in such a way that AOCB is a parallelogram. Find the value of ∠AOC. **

Solution:

**Given :**

Three points A, B and C lie on a circle with centre at O in such a way that AOCB is a parallelogram.

We know that sum of any two adjacent angles of a parallelogram is 180°.

∴ ∠AOC + ∠OAB = 180° ……(1)

Again, the central angle produced by the arc \(\overparen{A C}\) = reflex ∠AOC and angle in circle = ∠ABC

∴ by the theorem, reflex ∠AOC = 2 ∠ABC or, 360° – ∠AOC = 2 ∠AOC [∠ABC = ∠AOC, since opposite angles of any parallelogram are equal.]

or, 360° = 3 ∠AOC or, ∠AOC = \(\frac{360^{\circ}}{3}\) or, ∠AOC = 120°

∴ ∠AOC = 120°

**Example 4. The circumcentre of the isosceles triangle ABC is O and ∠ABC = 120°. It the radius of the circle be 5 cm, then determine the length of AB. **

Solution:

**Given :**

The circumcentre of the isosceles triangle ABC is O and ∠ABC = 120°. It the radius of the circle be 5 cm

Let us join O, A; O, B and O, C.

The central angle produced by the arc AC = reflex ∠AOC and angle in circle = ∠ABC = 120°

By theorem, reflex ∠AOC = 2 ∠ABC.

or, 360° – ∠AOC = 2 x 120°

or, 360° – ∠AOC = 240°

or, ∠AOC =360° – 240° = 120° ……..(1)

Again, in ΔABC, AB = BC [ABC is isosceles]

Now, in ΔOAB and ΔOCB, OA = OC [radii of same circle] A

B = BC and OB is common to both.

∴ ΔOAB = ΔOCB [by the condition of S-S-S congruent]

∴ ∠AOB = ∠BOC [similar angles of congruent triangles]

Now, ∠ABO + ∠CBO = ∠ABC = 120°

or, ∠ABO + ∠ABO = 120° [∠OBA = ∠OBC]

or, 2 ∠ABO = 120° or, ∠ABO = \(\frac{120^{\circ}}{2}\) = 60°

Again, ∠AOB + ∠COB = ∠AOC = 120° [by (1)]

or, ∠AOB + ∠AOB = 120° [∠COB = ∠AOB]

or, 2 ∠AOB = 120° or, ∠AOB = \(\frac{120^{\circ}}{2}\) =60°.

∴ in ΔAOB, ΔAOB = ∠ABO = 60°

∴ ΔAOB is equilateral.

∴ AB = AO = 5 cm [AO = radius = 5 cm]

Hence the required length of AB = 5 cm.

**Example 5. Two circles with centres A and B intersect each other at points C and D. The centre B of the other circle lies on the circle with centre A. If ∠CQD = 70°, then find the value of ∠CPD.**

Solution:

**Given :**

Two circles with centres A and B intersect each other at points C and D. The centre B of the other circle lies on the circle with centre A. If ∠CQD = 70°

Let us join A, C; B, C; A, D and B, D.

Now, in the circle with centre at B, angle in circle produced by the arc \(\overparen{C D}\) = ∠CQD and central angle = ∠CBD.

∴ by theorem, ∠CBD = 2 ∠CQD = 2 x 70° [∠CQD = 70°] =140°

Again, in the circle with centre A, the central angle produced by p the arc \(\overparen{C P D}\)= reflex ∠CAD and angle in circle = ∠CBD.

∴ by theorem, reflex ∠CAD = 2 ∠CBD

or, 360° – ∠CAD = 2. x 140° [∠CBD = 140°] or, ∠CAD = 360° – 280° = 80°

Again, in the circle with centre A, the central angle produced by the arc \(\overparen{C D}\) = ∠CAD and its angle in circle = ∠CPD.

By theorem, ∠CAD = 2 ∠CPD or, 80° – 2 ∠CPD [∠CAD = 80°]

or, ∠CPD = \(\frac{80^{\circ}}{2}\) = 40°

Hence the value of ∠CPD = 40°

## Solid Geometry Chapter 2 Theorems Related To Angles In A Circle

## Angles In A Circle Long Answer Type Questions

**Example 1. In the isosceles ΔABC, AB = AC. The circumcentre of ΔABC is O and the centre O lies on the opposite side of BC in which A lies. If ∠BOC = 100°, then find the value of ∠ABC and ∠ABO.**

Solution:

**Given :**

In the isosceles ΔABC, AB = AC. The circumcentre of ΔABC is O and the centre O lies on the opposite side of BC in which A lies. If ∠BOC = 100°,

In ΔBOC, OB = OC [radii of same circle]

∴ ∠OBC = ∠OCB

or, ∠OBC + ∠OBC = ∠OCB + ∠OBC

or, 2 ∠OBC = 180° – ∠BOC or, 2 ∠OBC = 180° – 100°

or, 2 ∠OBC = 80° or, ∠OBC = \(\frac{80^{\circ}}{2}\) = 40°.

Again, the central angle produced by the arc BC = ∠BOC and the angle in circle = ∠BAC.

By theorem, ∠BOC = 2 ∠BAC.

or ∠BAC = \(\frac{1}{2}\) ∠BOC = \(\frac{1}{2}\) x 100° =50°

Now, ∠ABC = 180° – (∠ACB + ∠BAC)

= 180° – ∠ABC – 50° [∠ACB = ∠ABC and ∠BAC = 50°]

= 130°-∠ABC

or, 2 ∠ABC = 130° or, ∠ABC = \(\frac{130^{\circ}}{2}\) = 65°

Then ∠ABO – ∠ABC – ∠OBC – 65° – 40° = 25°

Hence ∠ABC = 65 ° and ∠ABO = 25°

**Example 2. In the adjoining, if O is the centre of the circumcircle of ΔABC and ∠AOC = 110°, find the value of ∠ABC**

Solution:

**Given :**

In the adjoining, if O is the centre of the circumcircle of ΔABC and ∠AOC = 110°,

The angle in circle produced by the major arc \(\overparen{A B}\) = ∠ABC and the centre angle = reflex ∠AOC.

∴ by theorem, reflex ∠AOC = 2 ∠ABC or, 360° – ∠AOC = 2 ∠ABC

or, 360° – 110° = 2 ∠ABC [∠AOC = 110°]

or, 250° = 2 ∠ABC 250°

or, ∠ABC = \(\frac{250^{\circ}}{2}\) = 125°

∴ ∠ABC = 125°

**Example 3. O is the centre of the circle. If ∠AOD = 40° and ∠ACD = 35° then find the values of ∠BCO and ∠BOD.**

Solution:

**Given :**

O is the centre of the circle. If ∠AOD = 40° and ∠ACD = 35°

The central angle produced by the arc \(\overparen{A B}\) = ∠AOB and the angle in circle = ∠ACB = 35° (Given)

∴ by theorem, ∠AOB = 2 ∠ACB = 2 x 35° = 70°

Again, the central angle produced by \(\overparen{A D}\) = ∠AOD and angle in circle = ∠ACD or ∠ACO.

∴ by theorem, ∠AOD = 2 ∠ACO

or, 40° = 2 ∠ACO [∠AOD = 40° (Given)]

or, ∠ACO = \(\frac{40^{\circ}}{2}\) = 20°

Now, ∠BCO = ∠ACB = 35° + 20° = 55° [∠ACB = 35° (Given) and ∠ACO = 20°]

and ∠BOD = ∠AOD + ∠AOB = 40° + 70° = 110° [∠AOD = 40° (Given) and ∠AOB = 70°] ∠BCO = 55° and ∠BOD = 110°

**Example 4. Like the adjoining figure, draw two circles with centres C and D which intersect each other at the points A and B. Draw a straight line through A which intersects the circle with centre C at the point P and the circle with centre D at the point Q. **

**Prove that (1) ∠PBQ = ∠CAD; (2) ∠BPC = ∠BQD.**

Solution:

**Given :**

Like the adjoining figure, draw two circles with centres C and D which intersect each other at the points A and B.

In the circle with centre C, the central angle produced by the chord AP = ∠ACP and angle in circle = ∠ABP.

By theorem, ∠ACP = 2 ∠ABP or ∠ABP = \(\frac{1}{2}\) ∠ACP….(1)

Again, in the circle with centre at D, the central angle produced by the chord AQ = ∠ADQ and angle in circle ∠ABQ.

∴ by theorem, ∠ADQ = 2 ∠ABQ or, ∠ABQ = \(\frac{1}{2}\) ∠ADQ …(2)

Then, adding (1) and (2) we get,

∠ABP + ∠ABQ = \(\frac{1}{2}\) ∠ACP+ \(\frac{1}{2}\) ∠ADQ

or, ∠PBQ = \(\frac{1}{2}\)(∠ACP + ∠ADQ) = \(\frac{1}{2}\) (180° – 2 ∠PAC +180°- 2 ∠QAD)

[∠ACP + ∠PAC + ∠APC – 180°]

or, ∠ACP + ∠PAC + ∠PAC – 180° or, ∠ACP – 180° – 2 ∠PAC

Similarly, ∠ADQ = 180° – 2 ∠QAD

or, ∠PBQ = \(\frac{1}{2}\)[360° -2 (∠PAC + ∠QAD)] = \(\frac{1}{2}\) [360° -2 (180° – ∠CAD)]

= \(\frac{1}{2}\) [360° – 360° + 2 ∠CAD] = \(\frac{1}{2}\) x 2 ∠CAD = ∠CAD

∴ ∠PBQ = ∠CAD. [Proved (1)]

Now, let us join B, C and B, D

CP = CB (radii of same circle), ∠BPC = ∠PBC ……. (3)

BD = DQ (radii of same circle), ∠DBQ = ∠DQB……. (4)

Now, from (1) we get, ∠PBQ = ∠CAD.

or, ∠PBA + ∠QBA = ∠CAB + ∠DAB .

or, ∠PBC + ∠CBA + ∠DBA – ∠DBQ – ∠CAB + ∠DAB

or, ∠PBC + ∠CAB + ∠DAB – ∠DBQ = ∠CAB + ∠DAB

[CB = CA, ∴ ∠CBA = ∠CAB]

DB = DA, ∴ ∠DBA = ∠DAB

or, ∠PBC – ∠DBQ = 0 or, ∠PBC = ∠DBQ or, ∠BPC = ∠BQD [∠CBP = ∠BPC and ∠DBQ = ∠BQD]

∴ ∠BPC = ∠BQD. [Proved (2)]

Hence (1) ∠PBQ = ∠CAD and (2) ∠BPC = ∠BQD (Proved)

**Example 5. Each of two equal circles passes through the centre of the other and the two circles intersect each other at the points A and B. If a straight line through the point A intersects the two circles at points C and D, prove that ABCD is an equilateral triangle.**

Solution:

**Given :**

Each of two equal circles passes through the centre of the other and the two circles intersect each other at the points A and B. If a straight line through the point A intersects the two circles at points C and D

Let the two circles with centres at P and Q respectively are equal.

The two circles intersect each other at A and B.

The straight line CD passes through A and intersects the circle with centre P at C and with centre Q at D.

** To prove**: We have to prove that ΔBCD is equilateral.

** Construction**: Let us join P, Q; A, P; B, P; A, Q and B, Q. Let PQ intersects AB at S.

** Proof**: In triangles ΔAPS and ΔBPS, AP = BP [radii of same circle]

AS = BS [S is the mid-point of AB] and PS is common to both.

∴ ΔAPS ≅ ΔBPS, ∴ ∠APS = ∠BPS [similar angles of congruent triangles]……. (1)

Now, in the circle with centre at P, the angle in circle produced by the arc AB = ∠ACB and the central angle = ∠APB.

∴ ∠APB = 2 ∠ACB [by theorem]

or, ∠APS + ∠BPS = 2 ∠ACB

or, ∠APS + ∠APS = 2 ∠ACB

or, 2 ∠APS = 2 ∠ACB or, ∠APS = ∠ACB…… (2)

Also in the circle with centre at Q, the central angle produced by the arc \(\overparen{A D B}\) = reflex ∠AQB and angle in circle = ∠APB.

∴ by theorem, reflex ∠AQB = 2 ∠APB

or, 360° – ∠AQB – 2 ∠APB

or, 360° – ∠APB -2 ∠APB

[∠AQB = ∠APB,

Since ΔAPS = ΔAQS ⇒ ∠APS = ∠AQS,

Similarly, ∠BPS = ∠BQS]

or, 360° = 3 ∠APB or, ∠APB = \(\frac{360^{\circ}}{3}\) = 120°

or, 2 ∠APS =120° [∠APB = 2 ∠APS]

or, ∠APS = \(\frac{120^{\circ}}{2}\) = 60°

∴ ∠ACB = 60° [from (2)]

Similarly, it can be proved that ∠BDC = 60°

∴ the other angle of ΔBCD is 60°.

i.e., in mangle BCD, ∠BCD = ∠BDC = ∠CBD = 60° .

Hence ΔBCD is an equilateral triangle. (Proved).

**Example 6. S is the centre of the circumcircle of ΔABC and if AD ⊥ BC, prove that ∠BAD = ∠SAC.**

Solution:

**Given :**

S is the centre of the circumcircle of ΔABC and if AD ⊥ BC

S is the centre of the circum-circle of ΔABC,

∴ SA = SB = SC.

Also, AD ⊥ BC, ∠ADB = 90°, ∠ABD + ∠BAD = 90°

or, ∠ABD = 90° – ∠BAD……(1)

Now, in the circle with centre S, the central angle produced by the chord AC = ∠ASC and angle in circle = ∠ABC

∴ ∠ASC = 2 ∠ABC…… (2)

But SA = SC [radii of same circle]

∴ ∠SAC = ∠SCA….. (3)

Then, ∠ASC + ∠SAC + ∠SCA = 180°

or, ∠ASC + ∠SAC + ∠SAC = 180° [from (3)] .

or, ∠ASC + 2 ∠SAC = 180°

or, 2 ∠ABC + 2 ∠SAC – 180° [by (2)]

or, ∠ABC + ∠SAC = 90° [dividing by 2]

or, ∠ABD + ∠SAC = 90° [∠ABC and ∠ABD are same angle]

or, 90° – ∠BAD + ∠SAC = 90° [from (1)]

or, ∠SAC – ∠BAD = 0 or, ∠SAC = ∠BAD

Hence ∠BAD = ∠SAC (Proved)

**Example 7. Two chords AB and CD of a circle with centre O intersect each other at point P. Prove that ∠AOD + ∠BOC = 2 ∠BPC. If ∠AOD and ∠BOC are supplementary to each other, then prove that the two chords are perpendicular to each other.**

Solution:

**Given :**

Two chords AB and CD of a circle with centre O intersect each other at point P. Prove that ∠AOD + ∠BOC = 2 ∠BPC. If ∠AOD and ∠BOC are supplementary to each other

In the circle with centre at O, the two chords AB and CD intersect each other at P.

**To prove:** We have to prove that ∠AOD + ∠BOC = 2 ∠BPC and AB ⊥ CD when ∠AOD + ∠BOC = 180°.

** Construction**: Let us join O, A; O,B; O, C; O, D and B, D.

** Proof**: The central angle produced by the arc \(\overparen{\mathrm{AD}}\) = ∠AOD and angle in circle = ∠ABD.

∴ ∠AOD = 2 ∠ABD [by theorem] ……….(1)

Again, the central angle produced by the arc BC = ∠BOC and angle in circle = ∠BDC.

∴ ∠BOC = 2 ∠BDC (2) [by theorem]

Now, adding (1) and (2) we get, . ‘ .

∠AOD + ∠BOC = 2 ∠ABD + 2 ∠BDC = 2 (∠ABD + ∠BDC)…….(3)

But by producing DP of ΔPBD upto C, the produced external ∠BPC = internally opposite (∠PBD + ∠BDP) = internally opposite (∠ABD + ∠BDC)

[they are same angles]

∠ABD + ∠BDC = ∠BPC

∴ from (3) we get, ∠AOD + ∠BOC = 2 ∠BPC.

∴ ∠AOD + ∠BOC = 2 ∠BPC. (Proved)

Now, if ∠AOD + ∠BOC = 180°, then we get, 180° = 2 ∠BPC or, ∠BPC = = 90°

Hence AB ⊥ CD [∠BPC = 90°] [Proved]

**Example 8. If two chords AB and CD of a circle with centre O, when produced intersect each other at the point P, prove that ∠AOC – ∠BOD = 2 ∠BPC.**

Solution:

**Given :**

If two chords AB and CD of a circle with centre O, when produced intersect each other at the point P,

Let AB and CD be two chords of the circle with centre at O. Produced AB and CD intersect at P.

** To prove**: We have to prove that ∠AOC – ∠BOD = 2 ∠BPC

** Construction**: Let us join B, C.

** Proof**: The central angle produced by the arc \(\overparen{\mathrm{AC}}\) = ∠AOC and angle in circle = ∠ABC.

∠AOC = 2 ∠ABC…. (1) [by theorem]

Again, the central angle produced by the arc BD = ∠BOD and angle in circle = ∠BCD

∴ ∠BOD = 2 ∠BCD…… (2) [by theorem]

Now, subtracting (2) from (1) we get,

∠AOC – ∠BOD – 2 ∠ABC – 2 ∠BCD ….(3)

Again, by producing PB of ABPC we get, external ∠ABC.

∴ ∠ABC = internally opposite (∠BPC + ∠BCP)

or, ∠ABC = ∠BPC + ∠BCD.

or, 2 ∠ABC = 2 ∠BPC + 2 ∠BCD [multiplying by 2] or, 2 ∠ABC – 2 ∠BCD = 2 ∠BPC

∴ from (3) we get, ∠AOC – ∠BOD = 2 ∠BPC.

Hence, ∠AOC – ∠BOD = 2 ∠BPC. (Proved)

**Example 9. Draw a circle with the point A of quadrilateral ABCD as centre which passes through the ****points B, C and D. Prove that ∠CBD + ∠CDB = \(\frac{1}{2}\) ∠BAD.**

Solution:

Let a circle be drawn with centre at A of the quadrilateral ABCD, which passes through B, C and D.

** To prove**: We have to prove that ∠CBD + ∠CDB = \(\frac{1}{2}\)∠BAD.

** Construction**: Let us join A, B; B, C; C, D; D, A; B, D and C, A.

** Proof**: The central angle produced by the arc \(\overparen{\mathrm{BC}}\) = ∠BAC and angle in circle = ∠BDC .

∴ ∠BDC = \([\frac{1}{2}\)∠BAC…… (1) [by theorem]

Again, the central angle produced by \(\overparen{\mathrm{CD}}\) = ∠CAD and angle in circle = ∠CBD

∴ ∠CBD = \([\frac{1}{2}\) ∠CAD… ….(2) [by theorem]

Then, adding (1) and (2) we get,

∠BDC + ∠CBD = \(\frac{1}{2}\)∠BAC + \(\frac{1}{2}\) ∠CAD = \(\frac{1}{2}\) (∠BAC + ∠CAD) = \(\frac{1}{2}\) ∠BAD.

Hence ∠CBD + ∠CDB \(/frac{1}{2}\) = \(/frac{1}{2}\) ∠BAD.

**Example 10. O is the circumcentre of ΔABC and OD is perpendicular on the side BC prove that ∠BOD = ∠BAC**

Solution:

**Given :**

O is the circumcentre of ΔABC and OD is perpendicular on the side BC

The circumcentre of ΔABC is O and OD ⊥ BC.

**To prove:** We have to prove that ∠BOD = ∠BAC

**Construction:** Let us join O, A; O, B; and O, C.

**Proof:** In ΔBOD and ΔCOD, OB = OC [radii of same circle],

∠OBD = ∠OCD [OB = OC] and ∠BDO = ∠CDO [each is right angle]

∴ ΔBOD ≅ ΔCOD [by the condition of A-A-S congruence]

∴ ∠BOD = ∠COD [similar angles of congruent triangles]

∴ ∠BOC = ∠BOD + ∠COD = ∠BOD + ∠BOD [∠COD = ∠BOD] = 2 ∠BOD……. (1)

Now, the central angle produced by the chord BC = ∠BOC and angle in the circle = ∠BAC.

∴ ∠BOC = 2 ∠BAC [by theorem]……(2)

Then adding (1) and (2) we get,

2 ∠BOD = 2 ∠BAC or, ∠BOD = ∠BAC

Hence ∠BOD = ∠BAC (Proved)

## Solid Geometry Chapter 2 Theorems Related To Angles In A Circle

## Central Angle

We have seen earlier that the number of angles in circle produced by the same circular arc in the same circle are infinity.

Now the question arises is there any relation among these infinite number of angles in circle?

Let in a circle with centre at O, ∠ACB and ∠ADB are two angles in circle produced by the arc APB.

We shall see that if these two angles are measured by a protractor, then the two angles are equal.

Hence angles in the same segment of a circle are equal.

We shall now prove this theorem logically by the geometric method.

## Solid Geometry Chapter 2 Theorems Related To Angles In A Circle

## Central Angle Theorem

**Theorem: Prove that angles in the same segment of a circle are equal.**

** Given**: Let ∠ACB and ∠ADB are two angles in a circle with centre at O, produced by the circular arc \(\overparen{\mathrm{APB}}\)

** To prove**: ∠ACB = ∠ADB

** Construction**: Let us join O, A and O, B

** Proof**: ∠ACB is an angle in circle and ∠AOB is it’s central angle both produced by the same arc \(\overparen{\mathrm{APB}}\) of the circle.

∴ ∠AOB = 2 ∠ACB….. (1)

Similarly, ∠ADB is an angle in circle and ∠AOB is the central angle both produced by the circular arc APB of the circle.

∴ ∠AOB = 2 ∠ADB …… (2)

Now, from (1) and (2) we get, 2 ∠ACB = 2 ∠ADB, or, ∠ACB = ∠ADB

Hence angles in the same segment of a circle are equal (Proved).

## Solid Geometry Chapter 2 Theorems Related To Angles In A Circle

## Concylic Points

Three or more than three points are said to be concyclic if they lie on the same circle.

Such as, in the adjoining, the points A, B, C, and D lie on the circle with centre O. So, the points A, B, C, D are concyclic.

There are some properties and characteristics of concyclic points.

Such as, the opposite angles of the quadrilateral which is obtained by joining the four concyclic points are supplementary to each other.

Again, if a line segment joining two points subtends equal angles at two other points on the same side of it, then the four points are concyclic.

We shall now logically prove this theorem by the geometric method.

## Solid Geometry Chapter 2 Theorems Related To Angles In A Circle

## Concylic Points Theorem

**Theorem: ****Prove that if a line segment joining two points subtends equal angles at two other points on the same side of it, then four points are concyclic.**

** Given**: Let the line segment joining two points A and B subtend equal angles ∠ACB and ∠ADB, i.e., ∠ACB = ∠ADB.

** To prove**: We have to prove that A, B, C, D are concyclic.

** Construction**: Let us draw the circle passing through three non-collinear points A, B and C.

** Proof**: If the circle constructed above passes through the fourth point D, then the theorem is proved.

But if the circle does not pass through D, it will intersect AD or produced AD at a point.

Let the circle does not pass through D and intersect AD at E. Let us join E, B.

Now, ∠ACB = ∠AEB [angles in the same segment of a circle]

But given that ∠ACB = ∠ADB = ∠AEB = ∠ADB

But it is impossible since any external angle of a triangle can never be equal to its internally opposite angles.

So the circle passing through the three non-collinear points A, B, C must pass through point D.

Hence the four points A, B, C and D are concyclic. (Proved)

In the following examples, the various application of the above theorems are discussed thoroughly.

## Solid Geometry Chapter 2 Theorems Related To Angles In A Circle

## Concylic Points Multiple Choice Questions

**Example 1. In the adjoining figure, O is the centre of the circle; if ∠ACB = 30°, ∠ABC = 60°, ∠DAB = 35° and ∠DBC = x°, then the value of x is**

- 35
- 70
- 65
- 55

Solution: ∠DBC and ∠CAD are two angles in circle produced by the same arc \(\overparen{\mathrm{CD}}\) with centre at O.

∴ ∠DBC = ∠CAD……(1)

Now, in ΔABC, ∠ACB = 30° and ∠ABC = 60°

∴ ∠BAC = 180° – (∠ACB + ∠ABC)

= 180° – (30° + 60°) = 180° – 90° = 90°

Again, ∠DAB = 35° (Given),

∴ ∠CAD = ∠BAC – ∠DAB = 90° – 35° = 55°

∴ from (1) we get, ∠DBC = ∠CAD = 55°, x° = 55°

Hence x = 55

∴ 4. 55 is correct.

**Example 2. In the adjoining, O is the centre of the circle; if ∠B AD = 65°, ∠BDC = 45°, then the value of ∠CBD is**

- 65°
- 45°
- 40°
- 20°

Solution: ∠BAC = ∠BDC = 45° [∠BDC = 45° (Given)]

Again, ∠BAD = 65° (Given), ∠CAD = ∠BAD – ∠BAC = 65° – 45° = 20°

Now, both ∠CAD and ∠CBD are angles in circle produced by the same arc \(\overparen{\mathrm{CD}}\) of the circle.

∴ ∠CAD = ∠CBD

But ∠CAD = 20°, ∠CBD = 20°

∴ 4. 20° is correct.

**Example 3. In the adjoining, O is the centre of the circle. If ∠AEB = 110°, ∠CBE = 30°, then the value of ∠ADE is**

- 70°
- 60°
- 80°
- 90°

Solution: In the circle with centre at O, the angles in circle produced by the arc AB are ∠ADB and ∠ACB.

∴ ∠ADB = ∠ACB…….(1)

Now, if the side CE of ABCE be produced to A, then the external angle ∠AEB produced is equal to the internally opposite (∠CBE + ∠BCE),

i.e., ∠AEB = ∠CBE + ∠BCE

or, 110° = ∠CBE + ∠BCE [∠AEB =110° (Given)]

or, ∠BCE = 110° – ∠CBE or, ∠BCE = 110° – 30° [∠CBE = 30° (Given)]

or, ∠ACB = 80° [∠BCE and ∠ACB are same angles]

∴ ∠ADB = ∠ACB = 80° [from (1)] or, ∠ADB = 80°

∴ 3. 80° is correct.

**Example 4. In the adjoining, O is the centre of circle. If ∠BCD = 28°, ∠AEC = 38°, then the value of ∠AXB is **

- 56°
- 94°
- 38°
- 28°

Solution: In the circle with centre at O, the angles ∠BAD and ∠BCD are both angles in circle produced by the arc \(\overparen{\mathrm{BD}}\).

∴ ∠BAD = ∠BCD = 28° (Given)

Again, ∠AEC = 38°, ∴ ∠ADC = (∠AEC + ∠DAB)

[external ∠ADC = internally opposite (∠AEC + ∠DAE)]

∴ ADC = 38° + ∠BAD [∠DAB and ∠BAD are same angles] – 38° + 28° = 66°

Again, ∠AXB = internally opposite (∠XDC + ∠XCD)

= ∠ADC + ∠BCD = 66° + 28° = 94°

∴ 2. 94° is correct.

**Example 5. In the adjoining, O is the centre of the circle and AB is its a diameter. If AB ∥ CD, ∠ABC = 25°, then the value of ∠CED is**

- 80°
- 50°
- 25°
- 40°

Solution: Let us join B, D and A, D.

AB ∥ CD, ∴ ∠ABD + ∠CDB = 180°……..(1)

Again, in the circle with centre at O, ∠CED and ∠CBD are two angles in circle produced by the arc \(\overparen{\mathrm{AD}}\),

∴ ∠CED = ∠CBD…. (1)

Now, ∠ABC = 25° (Given), ∴ ∠BCD = 25° [AB ∥ CD, alternate angle]

Again, ∠ADC = ∠ABC [angles in the same segment of the circle]

AB is a diameter,

∴ ∠AOB is the central angle and angles in circle is ∠ADB both produced by the arc \(\overparen{\mathrm{AEB}}\).

∴ ∠ADB = \(\frac{1}{2}\)∠AOB = \(\frac{1}{2}\) x 180° = 90° [∴ ∠AOB is a straight angle]

Then ∠BDC = ∠ADC + ∠ADB = 25° + 90° = 115°

∴ in ΔBCD, ∠CBD = 180° – (∠BCD + ∠BDC) = 180° – (25° + 115°) = 180° – 140° = 40°

from (1) we get, ∠CED = ∠CBD = 40°

∴ 4. 40° is correct.

## Solid Geometry Chapter 2 Theorems Related To Angles In A Circle

## Concylic Points True Or False

**Example 1. In the adjoining AD and BE are perpendicular to BC and AC respectively of the ΔABC. The four points A, B, D, E are concyclic.**

Solution: True,

since the angles ∠AEB and ∠ADB are on the same side of AB.

Now, ∠AEB and ∠ADB are both right angles, ∠AEB = ∠ADB.

Hence the four points A, B, D, E are concyclic.

**Example 2. In ΔABC, AB = AC; BE and CF are respectively the bisectors of ∠ABC and ∠ACB and intersect the sides AC and AB at the points E and F respectively. Then the four points B, C, E, F are not concyclic.**

Solution: False

since in ΔABC, AB = AC.

∴ ∠ABC = ∠ACB or, \(\frac{1}{2}\)∠ABC = \(\frac{1}{2}\)∠ACB

or, \(\frac{1}{2}\)∠EBC = ∠BCF [BE and CF are the bisectors of ∠ABC and ∠ACB respectively.]

Now, in ΔBEC and ΔBFC we get,

∠EBC = ∠BCF, ∠ECB = ∠FBC and BC is common to both.

∴ ΔBEC ≅ ΔBFC [by the A-A-S condition of congruency]

∴ ∠BEC = ∠BFC.

But they are two such angles on the same side of BC at the points E and F that they are equal.

∴ B, C, E, F are concyclic.

But given that B, C, E, F are not concyclic.

Hence the given statement is false.

## Solid Geometry Chapter 2 Theorems Related To Angles In A Circle

## Concylic Points Fill In the Blanks

**Example 1. All angles in the same segment of a circle are _____**

Solution: Equal

**Example 2. If the line segment joining two points subtends equal angle at two other points on the same side, then the four points are ______**

Solution: Concyclic

**Example 3. If two angles on the circle formed by two arcs are equal, then the lengths of arcs are ______**

Solution: Equal.

## Solid Geometry Chapter 2 Theorems Related To Angles In A Circle

## Concylic Points Short Answer Type Questions

**Example 1. In the adjoining, O is the centre of the circle AC is the diameter and chord DE is ****parallel to the diameter AC. If ∠CBD = 60°, find the value of ∠CDE.**

Solution:

**Given :**

In the adjoining, O is the centre of the circle AC is the diameter and chord DE is parallel to the diameter AC. If ∠CBD = 60°,

Let us join D, O. Then the central angle produced by the chord CD is ∠COD. and the angle in circle by the same chord CD is ∠CBD.

∴ ∠COD = 2 ∠CBD or, ∠COD = 2 x 60° or, ∠COD =120°

∴ in ΔCOD, ∠OCD + ∠ODC = 180° – ∠COD = 180° – 120°

[the sum of three angles of a triangle is 180°]

∴ ∠OCD + ∠OCD = 60° [∠ODC = ∠OCD]

or, 2 ∠OCD = 60° or, ∠OCD = \(\frac{60^{\circ}}{2}\) = 30°

Again, AC | | DE, ∴ ∠ACD = ∠CDE [alternate angles]

or, ∠CDE = ∠OCD [∠ACD and ∠OCD are same angles]

Hence the value of ∠CDE – 30°.

**Example 2. In the adjoining, QS is the bisector of an angle ∠PQR, if ∠SQR = 35° and ∠PRQ = 32°, then find the value of ∠QSR.**

Solution:

**Given :**

In the adjoining, QS is the bisector of an angle ∠PQR, if ∠SQR = 35° and ∠PRQ = 32°,

Let us join O, Q and O, R. .

Now, ∠QOR is the central angle produced by the arc QR and ∠QPR and ∠QSR are two angles in circle produced by the same arc QR.

∴ ∠QOR = 2 ∠QPR ……. (1) and ∠QOR – 2 ∠QSR …… (2)

∴ from (1) and (2) we get, 2 ∠QPR = 2 ∠QSR or, ∠QPR = ∠QSR.

Now, ∠PQR = 2 ∠SQR [QS is the bisector of ∠PQR]

= 2 x 35° = 70° and ∠PRQ – 32° [Given]

∴ ∠QPR = 180° – (∠PQR + ∠PRQ)

= 180° – (70° + 32°) [∠PQR = 70° and ∠PRQ = 32°]

= 180° – 102° = 78°

Hence ∠QSR = 78°.

**Example 3. In the adjoining, O is the centre of the circle and AB is the diameter. If AB and CD are mutually perpendicular to each other and ∠ADC = 50°, then find the value of ∠CAD.**

Solution:

**Given :**

In the adjoining, O is the centre of the circle and AB is the diameter. If AB and CD are mutually perpendicular to each other and ∠ADC = 50°,

The angles in circle produced by the chord AC are ∠ADC and ∠ABC.

∴ ∠ABC = ∠ADC = 50°

Again, the central angle = ∠AOB and angle in circle = ∠ACB both produced by the arc \(\overparen{A D B}\).

∴ ∠AOB = 2 ∠ACB or, 180° = 2 ∠ACB [∠AOB = straight angle = 180°]

or, ∠ACB = \(\frac{180^{\circ}}{2}\) = 90°

Then ∠CAB = 180° – (∠ACB + ∠ABC) [sum of three angles of a triangle is 180°]

= 180° – (90° + 50°) = 180° – 140° = 40°

Again, AB and CD are perpendiculars to each other. Let AB and CD intersect each other at E.

∴ ∠EAD = 180° – (∠ADE + ∠AED)

= 180° – (50° + 90°) [∠ADE = ∠ADC = 50° and ∠AED = 90°]

= 180° – 140° = 40°

Now, ∠CAD = ∠EAC + ∠EAD

= 40° + 40° = 80° [∠EAC = ∠CAB = 40° and ∠EAD = 40°] .

∴ ∠CAD = 80°

**Example 4. In the adjoining, O is the centre of the circle and AB = AC; if ∠ABC = 32°, then find the value of ∠BDC. **

Solution:

**Given :**

In the adjoining, O is the centre of the circle and AB = AC; if ∠ABC = 32°,

In ΔABC, AB = AC; ∠ABC = ∠ACB = 32°

[∠ABC = 32°]

Now, ∠ACB and ∠ADB are angles in circle, produced by the chord AB.

∴ ∠ADB = ∠ACB = 32°….(1)

Again, ∠ABC and ∠ADC are two angles in circle, produced by the chord AC,

∴ ∠ADC = ∠ABC = 32°……(2)

Also, ∠BDC = ∠ADB + ∠ADC = 32° + 32° = 64° [from (1) and (2)]

∴ ∠BDC = 64°.

**Example 5. In the adjoining, BX and CY are the bisectors of ∠ABC and ∠ACB respectively. If AB = AC and BY = 4 cm, then find the length of AX. **

Solution:

**Given :**

In the adjoining, BX and CY are the bisectors of ∠ABC and ∠ACB respectively. If AB = AC and BY = 4 cm

In ΔABC, AB = AC, ∴ ∠ABC = ∠ACB

or, \(\)∠ABC = ∠ACB

or, ∠ABX = ∠BCY [BX and CY are the bisectors of ∠ABC and ∠ACB respectively.]

∴ by the converse theorem of theorem 35, arc AX = arc BY = 4 cm [BY = 4cm (Given)]

Hence the length of AX = 4cm.

## Solid Geometry Chapter 2 Theorems Related To Angles In A Circle

## Concylic Points Long Answer Type Questions

**Example 1. O is the orthocentre of ΔABC and AD ⊥ BC. If AD is produced it intersects the circumcircle of ΔABC at the point G. Prove that OD = DG.**

Solution:

**Given :**

O is the orthocentre of ΔABC and AD ⊥ BC. If AD is produced it intersects the circumcircle of ΔABC at the point G.

The perpendiculars AD, BE and CF on the sides BC, CA and AB respectively of ΔABC intersect each other at O.

So, O is the orthocentre of ΔABC. When AD is extended it intersects the circumcircle with centre P at point G.

** To prove**: We have to prove that OD = DG.

** Construction**: Let us join C, G.

** Proof**: ∠BAG and ∠BCG are two angles in circle, produced by the arc \(\overparen{B G}\).

∴ ∠BAG = ∠BCG.

or, 90° – ∠ABD = ∠BCG [AD ⊥ BC, ∴ ∠ADB = 90°]

or, 90° – (90° – ∠BCF) = ∠BCG [CF ⊥ AB, ∴ ∠BFC = 90°] or, ∠BCF = ∠BCG

or, ∠OCD = ∠GCD [∠BCF and ∠OCD also ∠BCG and ∠GCD are same angles]

∴ ∠OCD = ∠GCD …(1)

Now, in ΔCOD and ΔCGD,

∠CDO = ∠CDG [each is right angle],

∠OCD = ∠GCD [from (1)] and CD is common to both.

∴ ΔCOD ≅ ΔCGD [by the A-A-S condition of congruency]

∴ OD = DG [similar sides of congruent triangles]

Hence OD = DG (Proved)

**Example 2. I is the centre of the incircle of ΔABC; produced AI intersects the circumcircle of that triangle at the point P. Prove that PB = PC = PI.**

Solution:

**Given :**

I is the centre of the incircle of ΔABC; produced AI intersects the circumcircle of that triangle at the point P.

I is the incentre of ΔABC.

∴ ∠IBC = ∠IBA, ∠IAB = ∠IAC and ∠ICB = ∠ICA

∴ ∠IAB = ∠IAC, ∴ ∠PAB = ∠PAC

∴ PB = PC [by the converse theorem]…..(1)

Again, ∠BCP and ∠BAP are two angles in circle produced by the arc \(\overparen{P B}\).

∴ ∠BCP = ∠BAP

or, ∠PBC = ∠BAP [PB = PC, ∴ ∠BCP = ∠PBC]

Now, external ∠BIP of ΔAIB = internally opposite

(∠IAB + ∠IBA) = ∠IAB + ∠IBC

= ∠BAP + ∠IBC [∠IAB and ∠BAP are same angles]

= ∠PBC + ∠IBC

[∠PBC = ∠BAP] = ∠IBP

∴ ∠BIP + ∠IBP

∴ PB = BI or, [PB = PI, radius of same circle]……(2)

∴from (1) and (2) we get, PB = PC = PI. (Proved)

**Example 3. Ankita drew two circles which intersect each other at the points P and Q. Through the point P two straight lines are drawn so that they intersect one of the circles at the points A and B and the other circle at the points C and D respectively. Prove that ∠AQC = ∠BQD.**

Solution:

**Given :**

Ankita drew two circles which intersect each other at the points P and Q. Through the point P two straight lines are drawn so that they intersect one of the circles at the points A and B and the other circle at the points C and D respectively.

** Given**: Two circles with centres at O and O’ respectively intersect each other at the points P and Q.

Through the point P two straight lines APC and BPD are drawn which intersect the first circle at A and B and the second circle at C and D respectively.

** To prove**: We have to prove that ∠AQC = ∠BQD

** Construction**: Let us join A. Q; B, Q; C, Q and D, Q.

** Proof**: ∠APB and ∠AQB are two angles in circle produced by the arc AB in the circle with centre at O.

∴ ∠APB = ∠AQB ……. (1)

Again ∠CQD and ∠CPD are two angles in circle produced by the arc \(\overparen{C D}\) in the circle with centre at O’.

∴ ∠CQD = ∠CPD…..(2) .

But ∠APB = ∠CPD [opposite angles]

∴ ∠AQB = ∠CQD [from (1) and (2)]

Now, ∠AQD + ∠AQB = ∠AQD + ∠CQD [∠AQB = ∠CQD]

or, ∠BQD = ∠AQC

∴ ∠AQC = ∠BQD (Proved)

**Example 4. Two chords AB and CD of a circle are perpendicular to each other. If a perpendicular drawn to AD from the point of intersection of those two chords AB and CD is produced to meet BC at the point E, prove that the point E is the mid-point of BC. **

Solution:

**Given :**

Two chords AB and CD of a circle are perpendicular to each other. If a perpendicular drawn to AD from the point of intersection of those two chords AB and CD is produced to meet BC at the point E

AB ⊥ CD, ∴ ∠APD = ∠BPC = 90°

∴ ∠ADP = 90° – ∠DAP……. (1)

Now, ∠ADC and ∠ABC are two angles in circle, produced by the arc \(\overparen{A C}\).

∴ ∠ADC = ∠ABC…….(2)

Again, ∠BPE = ∠APF [opposite angles]…… (3)

Now, ∠APF – 90° – ∠PAF [∠AFP = 90°]

= ∠ADP [∠APD = 90°]……..(4)

From (3) and (4) we get, ∠BPE = ∠ADP = ∠ADC = ∠ABC = ∠PBE

∴ in ΔBPE, ∠BPE = ∠PBE. ∴BE = PE……….. (5)

Similarly, it can be proved that CE = PE…….(6)

Then from (5) and (6) we get, BE = CE

Hence E is the mid-point of BC. (Proved)

**Example 5. If in a cyclic quadrilateral ABCD, AB = DC, then prove that AC = BD.****[GP-X]**

Solution:

**Given :**

If in a cyclic quadrilateral ABCD, AB = DC

In the cyclic quadrilateral ABCD, AB = DC

∴ the angles in circle produced by the chords AB and CD are equal.

∴ ∠ACB = ∠ADB ……. (1) and

∠CBD = ∠CAD…….(2)

Moreover ACB = ∠CBD [AB = DC] in ΔEBC, ∠ECB = ∠EBC, ∴ BE = CE

Again, ∠ADB = ∠CAD [AB = DC]

∴ in ΔEAD, ∠EAD = ∠EDA ∴DE = AE

Now, AE + CE = DE + BE [AE = DE and CE = BE]

or, AC = BD (Proved)

**Example 6. OA is the radius of a circle with centre at O, AQ is its chord and C is any point on the circle. A circle passes through the points O, A, C intersects the chord AQ at the point P. Prove that CP = PQ.**

Solution:

**Given :**

OA is the radius of a circle with centre at O, AQ is its chord and C is any point on the circle. A circle passes through the points O, A, C intersects the chord AQ at the point P.

**Given:** OA is the radius of the circle with centre at O and AQ is its chord. C is any point on the circle.

A circle passes through the points O, A, C intersects the chord AQ at the point P. Let us join C, P.

**To prove:** We have to prove that CP= PQ

** Construction**: Let us join O, C; O, Q and O, P.

**Proof:** In the circle APOC, ∠OAP and ∠OCP are two angles in circle produced by the chord OP,

∴ ∠OAP = ∠OCP……. (1)

Again, in ΔOAQ, OA = OQ [radii of the same circle]

∴ ∠OAQ = ∠OQA……..(2)

from (1) and (2) we get, ∠OCP = ∠OQA …… (3)

∴ CP = PQ [in ΔCPQ, ∠QCP = ∠CQP]

Hence CP = PQ (Proved).

**Example 7. The triangle ABC is inscribed in a circle, the bisectors AX, BY and CZ of the angles ∠BAC, ∠ABC and ∠ACB intersect at the points X, Y, Z, on the circle respectively. Prove that AX is perpendicular to YZ. **

Solution:

**Given :**

The triangle ABC is inscribed in a circle, the bisectors AX, BY and CZ of the angles ∠BAC, ∠ABC and ∠ACB intersect at the points X, Y, Z, on the circle respectively.

Let ΔABC is inscribed in the circle with centre at O. AX, BY and CZ are the bisectors of ∠BAC, ∠ABC and ∠ACB respectively and intersect the circle at X, Y and Z respectively.

Let AX, BY and CZ intersect each other at O. Then O is the incentre of the circle. Again, let AX intersect YZ at the point E.

**To prove:** We have to prove that AX ⊥ YZ, i.e., AE ⊥ YZ

**Proof:** ∠ABY = ∠CBY (BY is the bisector of ∠ABC]

∴ arc AY = arc CY [by the converse]

∴ Similarly, arc BX = arc CX and arc AZ = arc BZ.

[angles in the same segment of a circle are equal.]

Again, arc AZ and arc BZ produced two equal central angles.

∴ ∠AOZ = ∠BOZ…..(1)

Similarly, ∠AOY = ∠COY [arc AY = arc CY]

or, ∠AOY = ∠BOZ [opposite angles] …… (2)

∴ ∠AOZ = ∠AOY or, ∠EOZ = ∠EOY …..(3)

Now, in ΔEOY and ΔEOZ, OY = OZ (radii of same circle),

∠EOY = ∠EOZ [from (3)] and OE is common to both.

∴ ΔEOY ≅ ΔEOZ [by the S-A-S condition of congruency]

∴ ∠OEY = ∠OEZ [similar angles of congruent triangles]

But these two angles are adjacent obtained when the line segment OE stands upon the line segment YZ and they are equal.

∴ each is right angle, i.e., ∠OEY = ∠OEZ = right angle.

∴ OE ⊥ YZ or AX ⊥ YZ.

∴ AX is perpendicular to YZ. (Proved)

**Class 10 Maths Wbbse Solutions**

**Example 8. ΔABC is inscribed in a circle,, the bisectors of the angles ∠BAC, ∠ABC and ∠ACB ****intersect at the points X, Y, Z on the circle respectively. Prove that in ΔXYZ, ∠YXZ = 90 – \(\frac{1}{2}\) ∠BAC**

Solution:

**Given :**

ΔABC is inscribed in a circle,, the bisectors of the angles ∠BAC, ∠ABC and ∠ACB intersect at the points X, Y, Z on the circle respectively.

** Given: **ΔABC is inscribed in the circle with centre at O.

The bisectors of ∠BAC, ∠ABC, and ∠ACB intersect the circle at the points X, Y and Z respectively. Let us join the points X, Y; Y, Z and Z, X.

** To prove**: ∠YXZ = 90° – \(\frac{1}{2}\)∠BAC.

** Proof**: ∠AXZ and ∠ACZ are two angles in circle produced by the arc AZ,

∴ ∠AXZ = ∠ACZ……..(1)

Similarly, ∠AXY and ∠ABY are two angles in circle produced by the arc AY.

∴ ∠AXY = ∠ABY…….. (2)

Now, adding (1) and (2) we get,

∠AXZ + ∠AXY = ∠ACZ + ∠ABY

or, ∠YXZ=\(\frac{1}{2}\)∠C+\(\frac{1}{2}\)∠B=\(\frac{1}{2}\) (∠C + ∠B)

= \(\frac{1}{2}\) (180° -∠A) [∠A + ∠B + ∠C = 180°]

= 90° –\(\frac{1}{2}\)∠A =90°-\(\frac{1}{2}\)∠BAC

Hence ∠YXZ = 90° –\(\frac{1}{2}\)∠BAC. [Proved]

**Class 10 Maths Wbbse Solutions**

**Example 9. The isosceles triangle ABC is inscribed in the circle with centre at O. If AP be a diameter passsing through A, then show that AP is the internal bisector of ∠BPC.**

Solution:** **

**Given :**

The isosceles triangle ABC is inscribed in the circle with centre at O. If AP be a diameter passsing through A

** Given**: ΔABC is an isosceles triangle of which AB = AC.

O is the centre of the circle in which ΔABC is inscribed and AP is a diameter of the circle.

** To prove**: AP is the internal bisector of ∠BPC, i.e., ∠APB = ∠APC.

** Construction**: Let us join the points O, B; O, C; B,P and C, P.

** Proof**: In ΔABC, AB = AC, ∴ ∠AOB = ∠AOC [equal segments of a circle produce equal front angles at the centre.]

or, 2 ∠APB = 2 ∠APC or, ∠APB = ∠APC.

Hence AP is the internal bisector of ∠BPC. (Proved)

**Example 10. In ΔABC, AB = AC and E is any point on the extended BC. The circumcircle of ΔABC intersects AE at the point D. Prove that ∠ACD = ∠AEC.**

Solution:

**Given :**

In ΔABC, AB = AC and E is any point on the extended BC. The circumcircle of ΔABC intersects AE at the point D.

**Given:** In ΔABC, AB = AC and E is any point on the produced BC.

The circumcircle of ΔABC intersects AE at point D.

**To prove:** We have to prove that ∠ACD = ∠AEC

** Construction**: Let us join B. D.

**Class 10 Maths Wbbse Solutions**

** Proof**: External ∠ACB of AΔACE = internally opposite (∠AEC + ∠CAE) = ∠AEC + ∠CBD [the angles produced by CD, ∠CAD = ∠CBD]

or, ∠ACB – ∠CBD = ∠AEC or, ∠ABC – ∠CBD = ∠AEC [AB = AC. ∴ ∠ACB = ∠ABC]

or, ∠ABD =∠AEC

or, ∠ACD = ∠AEC [the angles in circle produced by AD. ∠ABD = ∠ACD]

∴ ∠ACD = ∠AEC (Proved)

## Solid Geometry Chapter 2 Theorems Related To Angles In A Circle

## Angle In Semi Circle

In your previous classes you have learnt about what a semi-circle is. If a circle is divided into two equal parts by cutting it through any of its diameters, then each part of the two is called a semi-circle.

If we take any point P on the circumference of a semi-circle, then the angle obtained by joining two end points of the diameter of the semi-circle with the point P is called an angle in the semi-circle.

For example, in the following figure. AB is the diameter of the semi-circle and O is its centre. P_{1}, P_{2}, P_{3}, P_{4} are four points on its circumference.

If we join the endpoints A and B of the diameter AB with the points P_{1}, P_{2}, P_{3}, P_{4} respectively by some straight lines, we get the angles ∠AP_{1}B, ∠AP_{2}B, ∠AP_{3}B and ∠AP_{4}B.

Then these angles are called angles in a semi-circle. Now it is the question that what the value of these angles are and is there any relation among them.

If we measure these angles with hands-on trial, then we shall see that all these angles are equal and the value of each of them is 1 right angle, i.e..

∠AP_{1}B = ∠AP_{2}B = ∠AP_{3}B = ∠AP_{4}B= 1 right angle.

Hence angle in a semi-circle is a right angle.

We shall now prove this theorem logically by the method of geometry.

**Class 10 Maths Wbbse Solutions**

## Solid Geometry Chapter 2 Theorems Related To Angles In A Circle

## Angle In Semi Circle Theorem

**Theorem: Prove that angle in a semi-circle is a right angle.**

** Given**: ∠ACB is an angle in a semi-circle in a circle with centre at O.

** To prove **∠ACB = 1 right angle or 90°.

** Proof **: In the circle with centre at O, ∠AOB is the front angle and ∠ACB is the front angle in circle both produced by the same arc \(\overparen{A P B}\).

∴ ∠AOB = 2 ∠ACB ……. (1)

Now, ∠AOB = 1 straight angle [AOB is a line segment]

= 2 right angles or 180°.

∴ from (1) we get, 2 ∠ACB = 2 right angles or 180°^{ }or, ∠ACB = 1 right angle or 90°

∴ ∠ACB = 1 right angle or 90°.

Hence angle in a semi-circle is a right angle. (Proved)

In a similar way, by taking any angle in a semicircle we can prove that an angle in a semi-circle is a right angle.

We have discussed much more of the application of this theorem in the following examples.

## Solid Geometry Chapter 2 Theorems Related To Angles In A Circle

## Angle In Semi Circle Multiple Choice Questions

**Example 1. ****If PQ be a diameter of a circle with centre at O and if PR = RQ, then the value of ∠RPQ is**

- 30°
- 90°
- 60°
- 45°

Solution: PQ is a diameter of the circle with centre at O and R is any point on its circumference.

∴ ∠PRQ = 90° [by theorem]

Again, PR = RQ, ∴ ∠RPQ = ∠RQP…… (1) [opposite angles of equal sides of a triangle are equal]

Now, in ΔPQR, ∠PQR + ∠QRP + ∠ZRPQ = 180° [sum of three angles of a triangle is 180°]

or, ∠PQR + ∠RPQ + 90° = 180° [by (1) and ∠PRQ = 90°]

or, ∠RPQ + ∠RPQ + 90° = 180° [by (1) and ∠PRQ = 90°]

or, 2 ∠RPQ = 180° – 90° or, 2 ∠RPQ = 90°

or, ∠RPQ = \(\frac{90^{\circ}}{2}\)= 45° ∴ ∠RPQ = 45°

∴ 4. 45° is correct.

**Example 2. QR is a chord of the circle with centre at O and POR is a diameter of it. OD is perpendicular ****to QR. If OD = 4 cm, then the length of PQ is**

- 4 cm
- 2 cm
- 8 cm
- None of these

Solution: OD ⊥ QR, ∴ ∠ODR = 1 right angle

Again, POR is a diameter of the circle with centre at O.

∴ ∠PQR is an angle is semicircle.

∴ ∠PQR = 1 right angle.

∴OD ** ||** PQ and ∠PRQ is common to both ΔPQR and ΔODR. ΔPQR and ΔODR are similar triangles.

∴ \(\frac{O D}{P Q}=\frac{O R}{P R}\)

or, [/latex]\frac{4}{P Q}=\frac{O R}{2 O R}[because \mathrm{PR}=2 O R][/latex]

or, \(\frac{4}{P Q}=\frac{1}{2} \text { or, } P \mathrm{PQ}=8\)

∴ the length of PQ = 8 cm.

∴ 3. 8 cm is correct

**Aliter:** POR is a diameter of the circle with centre at O.

∴ ∠PQR is an angle in semicircle.

∴ ∠PQR = 1 right angle or 90°

∴ ΔPQR is a right-angled triangle of which PR is the hypotenuse.

∴ PR^{2} = PQ^{2} + QR^{2} …….(1) [by Pythagoras’ theorem]

Again, OD ⊥ QR, where is a chord of the circle.

∴ D is the mid-point of QR,

∴ DR = \(\frac{1}{2}\) QR.

∴ ΔODR is a right-angled triangle of which OR is the hypotenuse.

∴ OR^{2} = OD^{2} + DR^{2}

or, \(\left(\frac{1}{2} \mathrm{PR}\right)^2=4^2+\left(\frac{1}{2} \mathrm{QR}\right)^2\)

\(\cdot\left[\mathrm{OR}=\frac{1}{2} \mathrm{PR} \text { and } \mathrm{DR}=\frac{1}{2} \mathrm{QR} \text { and } \mathrm{OD}=4 \mathrm{~cm}\right]\)\(\frac{\mathrm{PR}^2}{4}=16+\frac{\mathrm{QR}^2}{4}\)

**Maths Solutions Class 10 Wbbse**

or, PR^{2} = 64 + QR^{2 }……(2)

Now, from (1) and (2), we get,

PQ^{2}+ QR^{2} = 64 + QR^{2 }

or, PQ^{2} = 64 or, PQ = √64 = 8.

∴ the length of PQ = 8 cm.

∴ 3. 8cm is correct.

**Example 3. AOB is a diameter of a circle. When the chords AC and BD are extended they meet at the point E. If ∠COD = 40°, then the value of ∠CED is **

- 40°
- 80°
- 20°
- 70°

Solution: Let us join B, C and C, D. AOB is the diameter of the circle With centre at O.

∴ ∠ACB is an angle in semicircle.

∴ ∠ACB = 90°…..(1)

Again, ∠BCE = 180° – ∠ACB = 180° – 90° [from (1)] = 90°…..(2)

Now, the front central angle produced by the chord CD is ∠COD = 40° and the front angle in circle is ∠CBD.

∴ ∠CBD * =* \(\frac{1}{2}\)∠COD

or,∠CBE = \(\frac{1}{2}\) x 40° = 20°…..(3) [∠CBD and ∠CBE are same angles]

So from ΔBCE we get, ∠BEC + ∠BCE + ∠CBE = 180°

or, ∠BEC + 90° + 20° = 180° [from (2) and (3)] or, ∠BEC = 180° – 110° = 70°

∴ ∠BEC = 70°

∴ 4. 70° is correct.

**Example 4. AOB is a diameter of a circle, If AC = 3 cm and BC = 4 cm, then the length of AB is**

- 3 cm
- 4 cm
- 5 cm
- 8 cm

Solution: AOB is a diameter of the circle with centre at O. ∠ACB is an angle in a semicircle

∴ ∠ACB = 90°

∴ from the right-angled triangle ABC by Pythagoras theorem we get,

AB^{2} = AC^{2} + BC^{2} = 3^{2} + 4^{2} = 25 [AC = 3 cm and BC = 4cm]

∴ AB = √25 = 5 the length of AB = 5cm

∴ 3. 5 cm is correct.

## Solid Geometry Chapter 2 Theorems Related To Angles In A Circle

## Angle In Semi Circle True Or False

**Example 1. The angle in circle produced by a minor axis is an obtuse angle.**

Solution: False since it will be an acute angle.

**Example 2. O is the mid-point of AB in ΔABC and OA = OB – OC; If a circle is drawn by taking AB as diameter, then the circle will pass through point C.**

Solution: True, since OA = OB = OC is also a radius of the circle with centre at O, so the circle will pass through the point C.

**Maths Solutions Class 10 Wbbse**

## Solid Geometry Chapter 2 Theorems Related To Angles In A Circle

## Angle In Semi Circle Fill In The Blanks

**Example 1. Semicircular angle is a ______ angle.**

Solution: Right

**Example 2. The angle in the segment of a circle which is less than a semi-circle is an _________ angle**

Solution: Obtuse

**Example 3. The circle drawn with hypotenuse of a right angled triangle as diameter passes through the _______**

Solution: Angular angle

## Solid Geometry Chapter 2 Theorems Related To Angles In A Circle

## Angle In Semi Circle Short Answer Type Questions

**Example 1. In the adjoining, O is the centre of the circle and AB is the diameter. If ∠BCE = 20°, ∠****CAE = 25°, then find the value of ∠AEC.**

Solution:

**Given :**

In the adjoining, O is the centre of the circle and AB is the diameter. If ∠BCE = 20°, ∠CAE = 25°

The angles in circle produced by the chord CA are ∠AEC and ∠ABC.

∴ ∠AEC = ∠ABC…… (1)

Again, the angles in circle produced by the chord BE are ∠BAE and ∠BCE.

∴ ∠BAE = ∠BCE – 20° [∠BCE = 20°]

Similarly, the angles in circle produced by the chord CE are ∠CBE and ∠CAE,

∴ ∠CBE = ∠CAE = 25° [∠CAE = 25°]

Again, ∠AEB is a semicircular angle, ∠AEB = 90°

Then, in ΔABE, ∠ABE = 180° – (∠BAE + ∠AEB) = 180° – (20°+ 90°) = 70°

∴ ∠ABC = ∠ABE – ∠CBE = 70°-25° = 45°

∴ from (1) we get, ∠AEC = ∠ABC = 45° ∴ ∠AEC = 45°

**Example 2. In isosceles triangle ABB = AC; if a circle is drawn with the side AB as diameter, then the circle intersects the side BC it the point D. When BD = 4 cm. find the length of CD.**

Solution:

**Given :**

In isosceles triangle ABB = AC; if a circle is drawn with the side AB as diameter, then the circle intersects the side BC it the point D. When BD = 4 cm.

Let O is the mid-point of AB.

The circle drawn with centre at O intersects the side BC at the point D.

Let us join A. D. Then ∠ADB is a semi-circular angle.

∴ ∠ADB = 90°, ∴ ∠ADC = 90°

Now in ΔABD and ΔACD, we have AB = AC, ∠ABD = ∠ACD and ∠ADB = ∠ADC.

∴ ΔABD = ΔACD [by die A-A-S condition of congruency ]

∴ BD = CD [similar sides of congruent triangles]

But BD = 4 cm (Given); CD = 4 cm.

**Example 3. Two chords AB and AC of a circle are perpendicular to each other. If AB = 4 cm and AC = 3cm, then find the radius of the circle. **

Solution:

**Given :**

Two chords AB and AC of a circle are perpendicular to each other. If AB = 4 cm and AC = 3cm,

AB and AC are perpendicular to each other.

∠BAC = 90°,

∴ ΔABC is a right-angled triangle of which BC is the hypotenuse.

∴ BC^{2 }= AB^{2} + AC^{2} [by Pythagoras theorem] = 4^{2}+ 3^{2} = 16 + 9 = 25

∴ BC = √25 = 5

Now, BC is one of the diameters of the circle.

∴ radius of the circle = \(\frac{5}{3}\) cm = 2.5 cm

**Example 4. Two chords PQ and PR of a circle are perpendicular to each other. If the radius of the circle be r cm. then find the length of the chord QR. **

Solution:

**Given :**

Two chords PQ and PR of a circle are perpendicular to each other. If the radius of the circle be r cm.

PQ and PR are perpendicular to each other, ∠QPR = 90°

∴ ∠QPR is a semicircular angle.

∴ QR is a diameter.

Since QR is a diameter of the circle,

radius of the circle = r cm (Given)

∴ QR = 2r cm.

**Maths Solutions Class 10 Wbbse**

**Example 5. AOB is a diameter of a circle. C is a point on the circle. If ∠OBC = 60°, then find the value of ∠OCA**

Solution:

**Given :**

AOB is a diameter of a circle. C is a point on the circle. If ∠OBC = 60°

AOB is a diameter of a circle.

C is a pont on circle

∴ ∠ACB = 1 right angle or 90°

Again, in ΔOBC, OB = OC (radii of same circle)

∴ ∠OCB = ∠OBC = 60° [∠OBC = 60° (Given)]

∴ ∠OCA= ∠ACB – ∠OCB OCB = 90° – 60° = 30°

∴ ∠OCA = 30°

** Example 6. In the adjoining, O is the centre of a circle and AB is one of its diameters. The length ****of the chord CD is equal to the radius of the circle. When AC and BD are extended, they intersect ****each other at a point P. Find the value of ∠APB.**

Solution:

**Given :**

In the adjoining, O is the centre of a circle and AB is one of its diameters. The length of the chord CD is equal to the radius of the circle. When AC and BD are extended, they intersect each other at a point P.

Let us join B, C.

Now, AB is the diameter of the circle,

∴ ∠ACB = 90° [semi-circular angle]

Again in ΔCOD, OC = OD = CD (Given)

∴ ΔCOD is an equilateral triangle.

∴ each angle of the triangle = 60° ∴ ∠COD = 60°

Now, the central angle produced by the chord CD is ∠COD and angle in circle is ∠CBD.

∴ ∠COD = 2 ∠CBD or, 60° = 2 ∠CBD or, ∠CBD = 30°

∴ in ΔPBC, ∠PCB = 90° [∠ACB = 90°], ∠PBC = ∠CBD = 30°.

∴ ∠BPC = 180° – (∠PCB + ∠PBC) = 180° – (90° + 30°) = 180° – 120° = 60°

∴ ∠APB = 60° [∠BPC = ∠APB]

## Solid Geometry Chapter 2 Theorems Related To Angles In A Circle

## Angle In Semi Circle Long Answer Type Questions

**Example 1. The angle B of the AABC is a right angle. If a circle is drawn with AC as diameter, then it intersects AB at a point D. Then which one of the following is correct?**

- AB > AD
- AB = AD
- AB < AD
- AB ≠ AD

Solution:

The angle B of ΔABC is a right angle.

∴ ∠ABC is a semicircular angle.

∴ AC is a diameter of the circle.

∴ The circle drawn with AC as diameter intersects AB at a point B, i.e., B and D are the same points.

∴ AB = AD

∴ 2. AB = AD is correct.

**Example 2. Prove that the circle drawn with any one of the equal sides of an isosceles triangle as diameter bisects the unequal side.**

Solution:

Let in ΔABC, AB = AC. The circle drawn with AB as diameter intersects BC at a point D.

**To prove:** D is the mid-point of BC, i.e., BD = CD.

** Proof:** The circle intersects BC at a point D.

∴ ∠ADB is a semi-circular angle.

∴ ∠ADB = 90° ⇒ ∠ADC = 90°

Again, in ΔABC, AB = AC

∴ ∠ABC = ∠ACB or, ∠ABD = ∠ACD

So, in ΔABD and ΔACD, ∠ADB = ∠ADC [each is 90°]

∠ABD = ∠ACD and AD is common to both.

ΔABD = ΔACD [by the A-A-S condition of congruency]

∴ BD = CD. [Similar sides of congruent triangle]

Hence D is the mid-point of BC, i.e., the circle bisects the unequal side. [Proved]

**Example 3. Parama drew two circles which intersect each other at the points P and Q. If the diameters of the two circles are PA and PB respectively, then prove that A, Q, B are collinear.**

Solution:

**Given :**

Parama drew two circles which intersect each other at the points P and Q. If the diameters of the two circles are PA and PB respectively,

Let two circles with centres O and O’ intersect each other at P and Q.

PA is a diameter of the circle with centre at O and PB is also a diameter of the circle with centre at O.

** To prove:** A, Q and B are collinear.

** Construction**: Let us join P, Q.

** Proof**: In the circle with centre at O, ∠AQP is a semi-circular angle. ∴ ∠PQA = 1 right angle.

Again, in the circle with centre at O’, ∠BQP is a semicircular angle. ∴ ∠PQB = 1 right angle.

Now, ∠PQA + ∠PQB = 1 right angle + 1 right angle.

or, ∠AQB = 2 right angle or, ∠AQB = 180° or, ∠AQB =1 straight angle.

∴ ∠AQB – 1 straight angle.

Hence the points A, Q, B are collinear. (Proved)

**Example 4. Debanjan drew a line segment PQ of which mid-point is R and two circles are drawn with PR and PQ as diameter. Debaratee drew a straight line through the point P which intersects the first circle at the point S and the second circle at the point T. Prove that PS = ST. **

Solution:

**Given :**

Debanjan drew a line segment PQ of which mid-point is R and two circles are drawn with PR and PQ as diameter. Debaratee drew a straight line through the point P which intersects the first circle at the point S and the second circle at the point T.

Let O be the centre of the circle drawn with PR as the diameter.

Let us join S,R and T, R. Since PR is a diameter. ∴ ∠PSR = 90° [Semicircular angle]

Obviously, ∠TSR = 90°, [Since PST is a straight line, so that ∠PST = 1^{ }straight line = 180°]

Now in ΔPSR and ΔTSR, PR = TR [radii of same circle with centre at R]

∠PSR = ∠TSR [ each is right angle] and SR is common to both.

∴ Δ PSR = Δ TSR [by the R-H-S condition of congruency]

∴ PS = ST [similar sides of two congruent triangles]

Hence PS = ST. (Proved)

**Example 5. Three points P, Q, R lie on a circle. The two perpendiculars PQ and PR at the point P intersect the circle at the points S and T respectively. Prove that RQ = ST.**

Solution:

**Given :**

Three points P, Q, R lie on a circle. The two perpendiculars PQ and PR at the point P intersect the circle at the points S and T respectively.

Three points P, Q, R are on the circle with centre at O. PS ⊥ PQ and PT ⊥ PR.

** To prove **RQ = ST.

** Construction: **Let us join the points Q, S; R, T; P, S and S, T.

** Proof: **PS ⊥ PQ, ∴ ∠QPS = 90°

∴ ∠QPS is a semi-circular angle.

∴ QS is a diametet, of which O is the mid point ∴ OQ = OS = radius

Similarly, PT ⊥ PR, ∴ ∠RPT = 90°

∴ ∠RPT is the semi-circular angle.

∴ RT is a diameter of which O is the mid-point.

∴ OR = OT = Radius.

Now, in Δ’s OQR and ΔOST, OQ = OS [radii of same circle]

∴ OR = OT [for similar reason] and ∠QOR = ∠SOT [opposite angles]

∴ ΔOQR ≅ ΔOST [by the S-A-S condition of congruency]

∴ QR = ST [similar sides of congruent triangles]

Hence RQ = ST. (Proved)

**Example 6. ABC is an acute-angled triangle. AP is the diameter of the circumcircle of the triangle ABC; BE and CF are perpendiculars on AC and AB respectively and they intersect each other at the point Q. Prove that BPCQ is a parallelogram. **

Solution:

**Given :**

ABC is an acute-angled triangle. AP is the diameter of the circumcircle of the triangle ABC; BE and CF are perpendiculars on AC and AB respectively and they intersect each other at the point Q.

ΔABC is an acute angled triangle. BE ⊥ AC and CF ⊥ AB and BE and CF intersect each other at a point Q.

AP is the diameter of the circumcircle of ΔABC.

**To prove:** BPCQ is a parallelogram.

**Construction:** Let us join O, C.

**Proof:** AP is a diameter of the circle. ∴ ∠ABP and ∠ACP are both semi circular angles.

∴ ∠ABP = ∠ACP = 90°……(1)

CF ⊥ AB, ∴ ∠CFB = 90°

∴ ∠FBQ = 90° – ∠FQB…….(2)

Again, BE⊥ AC, ∴ ∠BEC = 90°, ∴ ∠ECQ = 90° – ∠CQE . . . (3)

But ∠FQB = ∠CQE [Opposite angles]

∴ from (2) and (3) we get, ∠FBQ = ∠ECQ

Then ∠PBQ = ∠PBA – ∠FBQ = ∠ACP – ∠ECQ

[∠PBA =∠ACP and ∠FBQ = ∠ECQ] = ∠PCQ

∴ two opposite angles ∠PBQ and ∠PCQ of the quadrilateral BPCQ are equal.

Hence by the property of parallelogram, BPCQ is a parallelogram.

**Example 7. The internal and external bisectors of the vertical angle of a triangle intersect the circumcircle of the triangle at the points P and Q. Prove that PQ is a diameter of the circle. **

Solution:

The internal and external bisectors of the vertical angle of a triangle intersect the circumcircle of the triangle at the points P and Q.

**Given:** AP is the internal bisector of the vertical angle ∠A of the ΔABC and AE is the external bisector of the vertical ∠A of ΔABC and they intersect the circumcircle of the ΔABC at the points P and Q respectively.

**To prove:** PQ is the diameter of the circle.

**Proof:** ∠BAC + ∠CAD = 1 straight angle = 180° or, \(\frac{1}{2}\)∠BAC + \(\frac{1}{2}\) ∠CAD = 90° [dividing by 2]

or, ∠PAC + ∠CAE = 90° [AP is the internal bisector of ∠A and AE is the external bisector of ∠A]

or, ∠PAQ = 90°

∴ ∠PAQ is a semicircular angle.

Hence PQ is a diameter of the circle. (Proved)

**Example 8. AB and CD are two diameters of a circle. Prove that ADBC is a rectangle.**

Solution:

**Given :**

AB and CD are two diameters of a circle.

AOB and COD are two diameters of the circle with centre at O.

** To prove**: ADBC is a rectangle.

** Construction**: Let us join A, D; D, B; B, C and C, A.

** Proof**: AOB is a diameter. ∴ ∠ACB and ∠ADB are two semi-circular angles.

∠ACB = 1 right angle and ∠ADB = 1 right angle.

∴ Two opposite angles of the quadrilateral ADBC are both right angles.

Similarly, COD is a diameter. ∴ ∠CAD and ∠CBD are both semi-circular angles.

∴ ∠CAD = 1 right angle and ∠CBD = 1 right angle.

∴ Two other opposite angles of the quadrilateral ADBC are also right angles.

Again, in ΔAOD and ΔBOC, OA = OB, OD = OC [radii of same circle] and ∠COD = ∠BOC [opposite angles]

∴ ΔAOD ≅ ΔBOC [by the S-A-S condition of congruency]

∴ AD = BC [similar sides of congruent triangles] .

Similarly, it can be proved that AC = BD.

∴ opposite sides of the quadrilateral are equal and each of its angles is a right angle.

Hence ADBC is a rectangle. (Proved)

**Example 9. AB is diameter and AC is a chord of the circle with centre at O. The radius parallel to AC intersects the circle at D. Prove that D is the mid-point of the arc BC.**

** Solution**:

**Given :**

AB is diameter and AC is a chord of the circle with centre at O. The radius parallel to AC intersects the circle at D.

AB is a diameter of the circle with centre parallel to AC intersect the arc BC at the point D.

We have to prove that D is the mid-point of the arc BC.

** Construction**: Let us join O, C.

** Proof**: AC || OD and OC is the transversal of them,

∴ ∠OCA = ∠COD [alternate angle]…….. (1)

Again OA = OC [radii of same circle]

∴ ∠OCA = ∠OAC……..(2)

Now, ∠BOC = ∠BOD + ∠COD

or, internally opposite (∠OCA + ∠OAC) = ∠BOD + ∠COD

or, ∠COD + ∠OAC = ∠BOD + ∠COD [by (1)]

or, ∠OAC = ∠BOD or, ∠COD = ∠BOD [by (1)]

∴ ∠BOD = ∠COD, i.e. the two central angles produced by BD and CD are equal.

∴ arc BD = arc CD.

Hence D is the mid-point of the arc BC. (Proved)

**Example 10. Two circles intersect each other at the points P and Q. A straight line passing through P intersects one circle at A and the other circle at B. A straight line passing through Q intersect the first circle at C and the second circle at D. Prove that AC || BD.**

Solution:

**Given :**

Two circles intersect each other at the points P and Q. A straight line passing through P intersects one circle at A and the other circle at B. A straight line passing through Q intersect the first circle at C and the second circle at D.

Let two circles with centres at O and O’ intersect each other at the points P and Q respectively.

A straight line passing through P intersects the circle with centre O at the point A and the circle with centre O’ at the point B.

Another straight line passing through Q also intersects the first circle at C and the second circle at D.

We have to prove that AC || BD.

** Construction**: Let us join P, Q.

** Proof:** ΔPQC is a cyclic quadrilateral.

∴ ∠PAC + ∠PQC = 180° …….(1) [opposite angles of a cyclic quadrilateral are supplementary.]

Again, BPQD is also a cyclic quadrilateral.

∴ ∠PBD + ∠PQD = 180°……. (2) [for similar reason]

Now, adding (1) and (2) we get,

∠PAC + ∠PBD +∠PQC + ∠PQD = 360°

or, ∠PAC + ∠PBD + 180° = 360° [∠PQC + ∠PQD = 1 straight angle =180°] .

or, ∠PAC + ∠PBD = 360° – 180°

or, ∠PAC + ∠PBD = 180° .

i.e, the sum of two adjacent angles on the same side of the transversal AB of the two line segments AC and BD is 180°.

Hence AC || BD (Proved)

**Example 11. ABC is a cyclic equilateral triangle. If D be any point on the circular arc BC on the opposite side of the point A, then prove that DA = DB + DC**

Solution:

**Given :**

ABC is a cyclic equilateral triangle. If D be any point on the circular arc BC on the opposite side of the point A

Let ABC be a cyclic equilateral triangle inside the circle with centre at O.

D is any point on the circular arc BC on the opposite side of the point A.

We have to prove that DA = DB + DC.

** Construction:** Let us cut a part DE from DA equal to DC, i.e., DC = DE and let us join C, E.

**Proof:** In ΔCDE, DC = DE [as per construction]

∴ ∠DCE = ∠DEC …….(1)

∴ ΔABC is an equilateral triangle,

∠BAC – ∠ACB = ∠CBA = 60° and AB = BC = CA……..(2)

Now, two angles in circle produced by the arc AC are ∠ADC and ∠ABC

∴ ∠ADC = ∠ABC = 60° [by (2)]

∴ ∠CDE = 60°……(3)

∴ in ΔCDE, ∠CDE + ∠DCE + ∠DEC = 180° or, 60° + ∠DCE + ∠DCE = 180° [from (1) and (3)]

or, 2 ∠DCE = 180° – 60° or, ∠DCE = \(\frac{120^{\circ}}{2}\) = 60°.

∴ ∠DEC = 60°, i.e., each and every angle of ΔCDE is 60°, ΔCDE is equilateral.

∴ CD = DE = CE………(4)

Now, ∠ACE + ∠BCE = ∠ACB = 60° = ∠DCE = ∠BCD + ∠BCE

∴ ∠ACE = ∠BCD ……….(5)

Again, in Δ’s ACE and ΔBCD

∠CAE = ∠CBD [both are angles in circle produced by the same chord CD.]

∴ ∠ACE = ∠BCD [by (5)] and AC = BC [ΔABC is equilateral]

∴ ΔACE = ΔBCD [by the A-A-S condition of congruency]

∴ AE = BD [similar sides of congruent triangles]…….. (6)

Therefore, DA = DE + AE = DC + BD [DE = DC and AE = BD]

Hence DA = DB + DC (Proved).