WBBSE Solutions For Class 10 Maths Solid Geometry Chapter 3 Theorems Related To Cyclic Quadrilateral

WBBSE Solutions For Class 10 Maths Solid Geometry Chapter 3 Theorems Related To Cyclic Quadrilateral

Cyclic quadrilateral

If the four vertices of a quadrilateral entirely lie on the circumference of a circle, then it is called a cyclic quadrilateral.

For example, all four vertices of the following quadrilaterals ABCD. PQRS. DEFG. etc. entirely lie on the circumference of a circle. So these are arc all cyclic quadrilaterals.

Characteristics of cyclic quadrilateral 

1.  All the vertices of a cyclic quadrilateral always lie on the circumference of any circle.
2. The centre of the circle may be inside or outside the quadrilateral. Such as besides, the centre of the circle is inside the quadrilateral in the first and outside the quadrilateral in the second.
3. The opposite angles of any cyclic quadrilateral are supplementary.
4. Conversely, if two opposite angles of a quadrilateral arc are supplementary, i.e. if sum of the two opposite angles be 180°, then it is a cyclic quadrilateral.

The most important characteristic of a cyclic quadrilateral is that two of its opposite angles are supplementary to each other.

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

Solid Geometry Chapter 3 Theorems Related To Cyclic Quadrilateral

Cyclic Quadrilateral Theorem

Theorem 1. The opposite angles of a cyclic quadrilateral are supplementary.

 

Given: Let ABCD is a cyclic quadrilateral in the circle with centre at O.

To prove:

1. ∠ABC + ∠ADC = 2 right-angles
2. ∠BAD + ∠BCD = 2 right-angles

Construction: Let us join A, O and C, O.

Proof: The central angle produced by the circular arc \(\overparen{A B C}\) is ∠AOC and the angle in circle is ∠ADC.

∴ ∠AOC = 2 ∠ADC……. (1)

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

∴ reflex ∠AOC = 2 ∠ABC …..(2)

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

∠AOC + reflex ∠AOC = 2 ∠ADC + 2 ∠ABC

or, 4 right angles = 2 (∠ADC + ∠ABC)

or, ∠ADC + ∠ABC = \(\frac{4}{2}\) right angles = 2 right angles.

∴ ∠ABC + ∠ADC = 2 right angles ((1) proved)

Similarly, by joining O, B and O, D we can prove that ∠BAD + ∠BCD = 2 right angles.

Hence, 1. ∠ABC + ∠ADC = 2 right angles

2. ∠BAD + ∠BCD = 2 right angles (Proved)

From the above theorem, we can say conversely that if any two opposite angles of a quadrilateral be supplementary to each other, then it is a cyclic quadrilateral.

The question now arises whether this theorem is always true.

We shall now prove it logically by the method of geometry.

Theorem 2. If the opposite angles of any quadrilateral be supplementary, then the vertices of the quadrilateral are concyclic.

Given: Let PQRS be a quadrilateral in which two opposite angles ∠PQR and ∠PSR are supplementary

i.e., ∠PQR + ∠PSR = 2 right angles.

To prove: The vertices P, Q, R, and S of the quadrilateral are concyclic.

Construction: We know that through three non-collinear points P, Q, and R only one circle can be drawn.

Let us draw the circle. Also, let the constructed circle does not pass through S. The circle intersects PS or extended PS at the point T. Let us join T and R.

Proof: Given that ∠PQR + ∠PSR = 2 right angles.

But by construction, ∠PQR + ∠PTR = 2 right angles.

∴ ∠PQR + ∠PSR = ∠PQR + ∠PTR or, ∠PSR = ∠PTR.

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

∴ ∠PSR = ∠PTR is possible only if the points S and T coincide, i.e., if the circle is drawn through three points P, Q, and R passes through point S.

∴ the circle through the points P, Q, and R also passes through S.

Hence the points P, Q, R, and S are concyclic. (Proved)

Corollary Theorem: If one side of a cyclic quadrilateral is produced then the external angle thus obtained is equal to its internally opposite angle.

Given: Let ABCD be a cyclic quadrilateral. Side BC of it is extended to E so that an external angle ∠DCE is produced.

To prove ∠DCE = internally opposite ∠BAD.

Proof: ABCD is a cyclic quadrilateral,

∴ ∠BAD + ∠BCD = 180°……. (1)

Again, ∠BCD + ∠DCE = 1 straight angle =180°.

∴ ∠BAD + ∠BCD = ∠BCD + ∠DCE [from (1) and (2)] or, ∠BAD = ∠DCE

Hence ∠DCE = internally opposite ∠BAD (proved)

Solid Geometry Chapter 3 Theorems Related To Cyclic Quadrilateral

Cyclic Quadrilateral Multiple Choice Questions

“WBBSE Class 10 cyclic quadrilateral solved examples”

Example 1. In the adjoining, O is the centre of the circle and AB is one of its diameters. If ∠ADC = 120°, then the value of ∠BAC is

1. 50°
2. 60°
3. 30°
4. 40°

Solution:

Given:

In the adjoining, O is the centre of the circle and AB is one of its diameters. If ∠ADC = 120°,

ABCD is a cyclic quadrilateral ∠ABC + ∠ADC = 180° or, ∠ABC + 120° = 180° or, ∠ABC = 180° – 120° = 60°

Again, AOB is a diameter. ∠ACB is a semicircular angle, ∠ACB = 90°

Then in ΔABC, ∠BAC + ∠ABC + ∠ACB = 180°

or, ∠BAC + 60° + 90° = 180° [30°ABC = 60° and ∠ACB = 90°]

or, ∠BAC = 180° – 150° = 30°

∴ 3. 30° is correct.

The value of∠BAC is 3. 30°

Example 2. In the adjoining, O is the centre of the circle and AB is the diameter. ABCD is a cyclic quadrilateral. If ∠ABC = 65°, ∠DAC = 40°, then the value of ∠BCD is 

1. 75°
2. 105°
3. 115°
4. 80°

Solution:

Given:

In the adjoining, O is the centre of the circle and AB is the diameter. ABCD is a cyclic quadrilateral. If ∠ABC = 65°, ∠DAC = 40°

AB is the diameter of the circle with centre at O.

∴ ∠ACB is a semicircular angle. ∴ ∠ACB = 90°.

Now, ABCD is a cyclic quadrilateral,

∴ ∠ADC + ∠ABC = 180°

or, ∠ADC + 65° = 180° or, ∠ADC = 180° – 65° or, ∠ADC =115°

∴ in ΔACD, ∠ACD + ∠ADC + ∠CAD = 180°

or ∠ACD + 115° + 40° = 180° or, ∠ACD =180° – 155° or, ∠ACD = 25°

∴ ∠BCD = ∠ACB + ∠ACD = 90° + 25° = 115°

∴ 3. 115°  is correct.

The value of ∠BCD is  3. 115°

Example 3. In the adjoining, O is the centre of the circle and AB is one of its diameters. ABCD is a cyclic quadrilateral in which AB || DC and if ∠BAC = 25°, then the value of ∠DAC is

1. 50°
2. 25°
3. 130°
4. 40°

Solution:

Given:

In the adjoining, O is the centre of the circle and AB is one of its diameters. ABCD is a cyclic quadrilateral in which AB || DC and if ∠BAC = 25°,

AOB is a diameter of the circle,

∴ ACB is a semicircular angle. ∴ ∠ACB = 90°

Now, ∠ABC =180°- (∠BAC + ∠ACB) =180°- (25° + 90°) = 180° – 115° = 65°

Again, ABCD is a cyclic quadrilateral,

∴ ∠ADC + ∠ABC =180°

or, ∠ADC + 65° = 180° or, ∠ADC = 180° -65° =115°

Also, AB || DC and AC is their transversal,

∴ ∠ACD = alternate ∠BAC = 25°

∴ ∠DAC = 180° – (∠ADC + ∠ACD)

= 180° – (115° + 25°) = 180° – 140° = 40°

∴ 4. 40°  is correct.

“Theorems related to cyclic quadrilateral for Class 10 Maths”

Example 4. In the adjoining, ABCD is a cyclic quadrilateral. BA is produced to F. If AE || CD, ∠ABC = 92° and ∠FAE = 20°, then the value of ∠BCD is

1. 20°
2. 88°
3. 108°
4. 72°

Solution:

Given:

In the adjoining, ABCD is a cyclic quadrilateral. BA is produced to F. If AE || CD, ∠ABC = 92° and ∠FAE = 20°

ABCD is a cyclic quadrilateral.

∴ ∠ABC + ∠ADC = 180°

or, 92° + ∠ADC = 180° [∠ABC = 92°]

or, ∠ADC = 180° – 92° = 88°

Also given that AE || CD and AD is their transversal.

∴ ∠CDA = alternate ∠DAE

or, 88° = ∠DAE [∠CDA = 88°] ∴ ∠DAE = 88°

∴ ∠DAF = ∠DAE + ∠EAF = 88° + 20° [∠DAE = 88°] = 108°

∴ 3. 108° is correct.

Solid Geometry Chapter 3 Theorems Related To Cyclic Quadrilateral

Cyclic Quadrilateral True Or False

“Chapter 3 cyclic quadrilateral exercises WBBSE solutions”

Example 1. In the adjoining, AD and BE arc the perpendiculars on side BC and CA respectively of the ΔABC, Then, A, B, D, E are concyclic.

Solution:

Given:

In the adjoining, AD and BE arc the perpendiculars on side BC and CA respectively of the ΔABC

True; since the angles ∠ADB and ∠AEB on the same side of AB are equal for being right angles each.

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

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

Solution:

Given:

In ΔABC, AB = AC, BE and CF are the bisectors of the angles ∠ABC and ∠ACB and they intersect AC and AB at the points E and F respectively.

False; since in ΔABC, AB = AC,

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

⇒ ∠CBE = ∠BCF [BE and CF are the bisectors of ∠ABC and ∠ACB respectively]

⇒ ∠BEC = ∠BFC [∠ABC = ∠ACB]

Thus, the angles ∠BEC and ∠BFC on the same side of BC are equal, so the four points B, C, E, and F are concyclic.

But given that B, C, E, and F are not concyclic. Hence the statement is false.

Solid Geometry Chapter 3 Theorems Related To Cyclic Quadrilateral

Cyclic Quadrilateral Fill In The Blanks

Example 1. All angles in the same segment are ______

Solution: Equal

Example 2. If the line segment joining two points subtends equal angles 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 3 Theorems Related To Cyclic Quadrilateral

Cyclic Quadrilateral Short Answer Type Questions

Example 1. In the adjoining figure, if ∠BAD = 60°, ∠ABC = 80°, then find the values of ∠DPC and ∠BQC.

Solution:

Given:

In the adjoining figure, if ∠BAD = 60°, ∠ABC = 80°

∠DPC = ∠APB

= 180° – (∠PAB + ∠PBA)

= 180° – (60° + 80°) = 180° – 140° = 40°

Again, ABCD is a cyclic quadrilateral.

∴ ∠ADC + ∠ABC =180°

or, ∠ADC + 80° = 180° or, ∠ADC = 180° – 80° = 100°

∴ ∠QDA =100°

∴ ∠BQC = ∠AQD

= 180° – (∠QAD + ∠QDA) = 180° – (60° + 100°) [∠QAD = 60° (given), ∠QDA = 100°]

= 180° – 160° = 20°

∴ ∠DPC = 40° and ∠BQC = 20°

“Class 10 Maths properties of cyclic quadrilaterals”

Example 2. In the adjoining, two circles with centres P and Q intersect each other at points B and C, ACD is a line segment. If ∠ARB = 150°, ∠BQD =x°, then find the value of x.

Solution:

Given:

In the adjoining, two circles with centres P and Q intersect each other at points B and C, ACD is a line segment. If ∠ARB = 150°, ∠BQD =x°

ARBC is a cyclic quadrilateral,

∴ ∠ARB + ∠ACB = 180°

or, 150° + ∠ACB = 180° or, ∠ACB = 180° – 150° = 30°

Again, ∠ACD = ∠ACB + ∠BCD = 1 straight angle = 180°

or, 30° + ∠BCD = 180° or, ∠BCD = 180° – 30° = 150°

Now, in the circle with centre Q, ∠BCD is an angle in circle and reflex ∠BQD is its central angle both produced by the circular arc \(\overparen{B D}\)

∴ reflex ∠BQD = 2 ∠BCD

or, 360° – ∠BQD = 2 ∠BCD or, 360° – x° = 2 x 150° or, x° = 360° – 300° or, x° = 60°

Hence the value of x is 60.

Example 3. In the adjoining, two circles intersect each other at the points P and Q. If ∠QAD = 80° and ∠PDA = 84°, then find the value of ∠QBC and ∠BCP.

Solution:

Given:

In the adjoining, two circles intersect each other at the points P and Q. If ∠QAD = 80° and ∠PDA = 84°,

In the circle with centre at O, ADPQ is a cyclic quadrilateral.

∴ ∠ADP + ∠AQP = 180° or, 84° + ∠AQP = 180° [∠ADP = 84°]

or, ∠AQP – 180° – 84° = 96°

Now, ∠BQP + ∠AQP = 1 straight angle or, ∠BQP + 96° = 180° [∠AQP = 96°] or, ∠BQP = 180° – 96° = 84°

Again, BQPC is a cyclic quadrilateral.

∴ ∠BCP + ∠BQP = 180° or, ∠BCP + 84° = 180°

or, ∠BCP = 180° – 84° = 96°

Now, ∠DPQ + ∠DAQ = 180° or, ∠DPQ + 80° =180° or, ∠DPQ = 180° – 80° = 100°

∴ ∠DPQ + ∠QPC = 1 straight angle = 180°

or, 100° + ∠QPC = 180° or, ∠QPC = 180° – 100° = 80°

Again, ∠QPC + ∠QBC = 180° [BQPC is a cyclic quadrilateral]

or, 80° + ∠QBC = 180° or, ∠QBC = 180° – 80° = 100°.

Hence ∠QBC = 100° and ∠BCP = 96°.

Example 4. In the adjoining, O is the centre of the circle and AC is its diameter of it. If DC || EB, ∠AOB = 80° and ∠ACE = 10°, then find the value of ∠BED.

Solution:

Given:

In the adjoining, O is the centre of the circle and AC is its diameter of it. If DC || EB, ∠AOB = 80° and ∠ACE = 10°,

Given that DC || EB and CE is their transversal.

∴ ∠DCE = ∠BEC [alternate angle]……. (1)

Now, ∠AOB = 2 ∠ACB [central angle is twice of angle in circle]

or, 80° = 2 ∠ACB [∠AOB = 80° given]

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

Also, ∠BOC = 180° – ∠AOB = 180° – 80° = 100°

Again, the central angle produced by the chord \(\overparen{\mathrm{BC}}\) is ∠BOC and angle in circle is ∠BEC.

∴ ∠BEC= \(\frac{1}{2}\) ∠BOC = 100° =50° [∠BOC = 100°]

∴ ∠ECD = 50° [by (1)]

Also, ∠BED + ∠BCD =180° [BCDE is a cyclic quadrilateral]

or, ∠BED = 180° – ∠BCD =180° – 100° = 80°

[∠BCD = ∠ACB + ∠ACE + ∠ECD= 40° + 10° + 50° = 100°] ∴ ∠BED = 80°.

“Understanding cyclic quadrilaterals in Class 10 Maths”

Example 5. In the adjoining, O is the centre of the circle and AB is a diameter. If ∠AOD = 140° and ∠CAB = 50°, then find the value of ∠BED.

Solution:

Given:

In the adjoining, O is the centre of the circle and AB is a diameter. If ∠AOD = 140° and ∠CAB = 50°,

Given that ∠AOD = 140°

∴ reflex ∠AOD = 360° – 140° = 220°

Again, the central angle produced by the arc \(\overparen{\mathrm{ABD}}\) is reflex ∠AOD and angle in circle = ∠ACD.

∴ reflex ∠AOD = 2 ∠ACD

or, 220° = 2 ∠ACD or, ∠ACD = \(\frac{220^{\circ}}{2}\) = 110°

Given that ∠CAB = 50°

Now, in ΔACE, ∠AEC + ∠EAC + ∠ACE = 180° or, ∠AEC = 180° – 50° – 110° = 20°

Hence ∠BED = 20°

Solid Geometry Chapter 3 Theorems Related To Cyclic Quadrilateral

Cyclic Quadrilateral Long Answer Type Questions

Example 1. In the adjoining the diagonals of the cyclic quadrilateral PQRS intersect each other at the point X in such a way that ∠PRS = 65° and ∠RQS = 45°. Find the values of ∠SQP and ∠RSP

Solution:

Given:

In the adjoining the diagonals of the cyclic quadrilateral PQRS intersect each other at the point X in such a way that ∠PRS = 65° and ∠RQS = 45°.

Two angles in circle produced by the arc PS are ∠PRS and ∠PQS.

∴ ∠PQS = ∠PRS – 65° ∴ ∠SQP = 65°

Again, two angles in circle produced by the arc RS are ∠RQS and ∠RPS.

∴ ∠RQS = ∠RPS ⇒∠RPS = 45° [ ∠RQS = 45°]

Then in ΔPRS, ∠RSP + ∠PRS + ∠RPS = 180° or, ∠RSP + 65° + 45° = 180° [ ∠RPS = 45°] or, ∠RSP = 180° – 110° = 70°

∴ ∠SQP = 65° and ∠RSP = 70°

“Step-by-step solutions for cyclic quadrilateral problems Class 10”

Example 2. The side AB of the cyclic-quadrilateral ABCD is extended to X. If ∠XBC = 82° and ∠ADB = 47°. Find the value of ∠BAC.

Solution:

Given:

The side AB of the cyclic-quadrilateral ABCD is extended to X. If ∠XBC = 82° and ∠ADB = 47°.

Given that ∠XBC = 82°

∴ ∠ABC = 180° – ∠XBC = 180° – 82° = 98°

Again, two angles in circle produced by the arc AB are ∠ADB and ∠ACB, ∠ADB = ∠ACB, ∠ACB = 47°.

Now in ΔABC, ∠BAC + ∠ACB + ∠ABC = 180°

or, ∠BAC + 47° + 98° = 180° or, ∠BAC = 180° – 145° = 35°

Hence ∠BAC = 35°

Example 3. Two sides PQ and SR of the cyclic quadrilateral when produced meet at the point T. O is the centre of the circle; If ∠POQ = 110°, ∠QOR = 60°, ∠ROS = 80°, then find the value of ∠RQS and ∠QTR.

Solution:

Given:

Two sides PQ and SR of the cyclic quadrilateral when produced meet at the point T. O is the centre of the circle; If ∠POQ = 110°, ∠QOR = 60°, ∠ROS = 80°,

The angle in circle produced by the arc RS is ∠RQS and the corresponding central angle is ∠ROS = 80° (Given).

∴ ∠RQS = \(\frac{1}{2}\) ∠ROS = \(\frac{1}{2}\) x 80° = 40°.

Now, ∠ROP = ∠ROQ + ∠POQ = 60° + 110° [∠QOR = 60° and ∠POQ =110°] ∠ROP =170° .

Then, ∠RSP = \(\frac{1}{2}\)∠ROP [the angle in circle is ∠RSP and central angle is ∠ROP produced by the arc \(\overparen{\mathrm{RQP}}\)

= \(\frac{1}{2}\) x 170° = 85°

Again, ∠SOQ = ∠QOR + ∠SOR = 60° + 80° = 140°

Then ∠SPQ = \(\frac{1}{2}\)∠SOQ [angle in circle is ∠SPQ and central angle is ∠SOQ produced by the arc \(\overparen{\mathrm{SRQ}}\).

= \(\frac{1}{2}\) x 140° = 70°

∴ in Δ PTS, ∠PTS + ∠SPT + ∠PST = 180° .

or, ∠PTS + 70° + 85° = 180° [∠SPQ = 70° and ∠RSP = 85°] or, ∠PTS = 180° – 70° – 85° = 25° ∴ ∠QTR = 25°

∴ ∠RQS = 40° and ∠QTR = 25°

Example 4. Two circles intersect each other at the points P and Q. Two straight lines through P and Q intersect one circle at the points A and C and the other circle at B and D. Prove that AC 11 BD.

Solution:

Given:

Two circles intersect each other at the points P and Q. Two straight lines through P and Q intersect one circle at the points A and C and the other circle at B and D.

We know that two opposite angles of a cyclic quadrilateral are supplementary.

∴ for the quadrilateral ACQP,

∠CAP + ∠PQC = 180°……(1) and for the quadrilateral BPQD

∠PBD +∠PQD =180°…… (2)

Now adding (1) and (2), we get, ∠CAP + ∠PBD + ∠PQC + ∠PQD = 360°

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

or, ∠CAP + ∠PBD +180° = 360° or, ∠CAP + ∠PBD = 360° – 180° or, ∠CAP + ∠PBD =180°

But the common transversal of the straight lines AC and BD is AB and two adjacent angles on the same side of AB are ∠CAB and ∠DBA, the sum of which is 180°.

AC || BD [the sum of two adjacent angles on the same side of the common transversal of two straight lines is 180°]

Hence AC || BD. (Proved)

“WBBSE Mensuration Chapter 3 practice questions on cyclic quadrilaterals”

Example 5. Two straight lines are drawn through any point X exterior to a circle to intersect the circle at points A, B and points C, D respectively. Prove that ΔXAC and ΔXBD are equiangular.

Solution:

Given:

Two straight lines are drawn through any point X exterior to a circle to intersect the circle at points A, B and points C, D respectively.

 

ABCD is a cyclic quadrilateral.

∴ ∠ABD + ∠ACD =180° …… (1)

Again, ∠XCA + ∠ACD = 1 straight angle = 180°…..(2)

Then, by subtracting (1) from (2) we get,

∠XCA – ∠ABD = 0 or, ∠XCA = ∠ABD.

Again in cyclic quadrilateral ABDC, ∠BAC + ∠BDC = 180°……(3)

and ∠XAC + ∠BAC = 1 straight angle =180°……(4)

Then subtracting (3) from (4) we get,

∠XAC – ∠BDC = 0 or, ∠XAC = ∠BDC

∴ in triangles ΔXAC and ΔXBD,

∠XCA = ∠XBD [∠ABD = ∠XBD] and ∠XAC = ∠XDB

∴ two angles of each of ΔXAC and ΔXBD are equal.

Hence ΔXAC and ΔXBD are equiangular. (Proved)

Example 6. Two circles intersect each other at the points G and H. A straight line is drawn through the point G which intersect two circles at points P and Q and the straight line through the point H parallel to PQ intersects the two circles at the points R and S. Prove that PQ = RS. 

Solution:

Given:

Two circles intersect each other at the points G and H. A straight line is drawn through the point G which intersect two circles at points P and Q and the straight line through the point H parallel to PQ intersects the two circles at the points R and S.

Let us join P, R and Q, S.

As per the question, PQ || RS and PS is their transversal.

∴ ∠SPQ = ∠RSP…… (1) [alternate angles]

Similarly, PQ || RS and GH is their transversal.

∴ ∠PGH = ∠SHG ……(2) [alternate angles]

Again, PRHG is a cyclic quadrilateral

∴ ∠PGH + ∠PRH = 180° ….(3)

Similarly, QSHG is a cyclic quadrilateral.

∴ ∠SQG + ∠SHG =180°……(4)

From (3) and (4) we get, ∠PGH + ∠PRH = ∠SQG + ∠SHG or, ∠PRH = ∠SQG [from (2)] or, ∠PRS = ∠SQP…..(5)

Now, in ΔPRS and ΔPQS, ∠PSR = ∠QPS [from (1)] ∠PRS = ∠SQP [from (5)] and PS is common to both.

∴ ΔPRS = ΔPQS [by the A-A-S condition of congruency]

∴ RS = PQ [similar sides of congruent triangles]

Hence PQ = RS (proved)

Example 7. In triangle ABC, AB = AC and E is any point on the extended BC. If the circumcircle of ΔABC intersect AE at the point D, then prove that ∠ACD = ∠AEC.

Solution:

Given:

In triangle ABC, AB = AC and E is any point on the extended BC. If the circumcircle of ΔABC intersect AE at the point D,

In ΔABC, AB = AC,

∴ ∠ABC = ∠ACB………..(1)

Now, ABCD is a cyclic quadrilateral,

∴ ∠ABC + ∠ADC = 180°

or, ∠ACB + ∠ADC = 180° …… (2) [from (1)]

Again, ∠ACB + ∠ACE = 1 straight angle = 180°…….(3)

From (2) and (3) we get, ∠ACB + ∠ADC = ∠ACB + ∠ACE

or, ∠ADC = ∠ACE or, ∠DCE + ∠DEC = ∠ACD + ∠DCE or, ∠DEC = ∠ACD

∴ ∠ACD = ∠DEC.

Hence ∠ACD = ∠AEC (Proved)

Example 8. ABCD is a cyclic quadrilateral. The chord DE is the external bisector of ZBDC. Prove that AE (or extended AE), is the external bisector of ∠BAC.

Solution:

Given:

ABCD is a cyclic quadrilateral. The chord DE is the external bisector of ZBDC.

Two angles in circle produced by the arc BC are ∠BAC and ∠BDC.

∴ ∠BAC = ∠BDC……(1)

Again, two angles in circle produced by the are CF arc ∠CAE and ∠CDE.

∴ ∠CAE = ∠CDE….(2)

Now, ∠FDE + ∠CDE + ∠BDC = 1 straight angle = 180°.

or, ∠CDE + ∠CDE + ∠BDC = 180° [DE is bisector of ∠FDC  ∴ ∠FDE = ∠CDE]

or, 2 ∠CDE + ∠BDC = 180°…… (3)

Again, ∠GAE + ∠CAE + ∠BAC = 1 straight angle = 180°…… (4) .

From (3) and (4) we get, 2 ∠CDE + ∠BDC = ∠GAE + ∠CAE + ∠BAC

or, 2 ∠CDE = ∠GAE + ∠CAE [from (1)]

or, 2 ∠CAE = ∠GAC

or, ∠CAE = \(\frac{1}{2}\) ∠GAC

∴ AE is the bisector of ∠GAC.

Hence AE is the external bisector of ∠BAC. (Proved)

“Examples of cyclic quadrilateral calculations for Class 10”

Example 9. BE and CF are perpendicular on sides AC and AB of triangle ABC respectively. Prove that four points B, C, E, F are concyclic. Also prove that two angles of each of ΔAEF and ΔABC are equal.

Solution:

Given:

BE and CF are perpendicular on sides AC and AB of triangle ABC respectively.

BE ⊥ AC, ∠BEC = 90°

Also, CF ⊥ AB, ∠BFC = 90°, i.e., at two points E and F on the same side of BC, the two produced angles are equal.

∴ B, C, E, F are concyclic. (Proved)

Now, in ΔAEF, ∠AEF = ∠AEB – ∠BEF

= 90° – ∠BEF [∠AEB = 90°]

= 90° – ∠BCF [∠BEF = ∠BCF same angles in circle produced by the arc BF]

= ∠CBF [∠BFC = 90°]

Again, ∠AFE = ∠AFC – ∠CFE = 90° – ∠CFE [∠AFC = 90°]

= 90° – ∠CBE [∠CFE = ∠CBE, same angle in circle produced by the arc CE] = ∠BCE [∠BEC = 90°]

∴ in Δ’s AEF and ΔABC, ∠AEF = ∠ABC [∠CBF = ∠ABC] and ∠AFE = ∠ACB [∠BCE = ∠ACB]

Hence two angles of each of ΔAEF and ΔABC are equal. (Proved).

Example 10. ABCD is a parallelogram. A circle passing through points A and B intersect the sides AD and BC at points E and F respectively. Prove that the four points E, F, C, and D are concyclic.

Solution:

Given:

ABCD is a parallelogram. A circle passing through points A and B intersect the sides AD and BC at points E and F respectively.

Given: ABCD is a parallelogram. A circle with centre at O passing through the points A and B intersect the sides AD and BC at the points E and F respectively.

To prove: E, F, C, and D are concyclic.

Proof: As per ∠AEF + ∠DEF = 1 straight angle = 180°…..(1)

Again, ∠ABC + ∠BCD = 180° adjacent angles of a parallelogram]

[ABFE is a cyclic quadrilateral, ∴ ∠ABf + ∠AEF = 180° or, ∠ABF = 180° – ∠AEF or, ∠ABC – 180° – ∠AEF]

or, 180° – ∠AEF + ∠BCD = 180° or, ∠DEF + ∠BCD = 180° [180° – ∠AEF = ∠DEF] or, ∠DEF + ∠FCD = 180°

i.e.., two opposite angles of the quadrilateral EFCD are supplementary to each other.

Hence E, F, C, and D are concyclic. (Proved)

Example 11. ABCD is a cyclic quadrilateral. The two sides AB and DC are produced to meet in the point P and other two sides AD and BC are produced to meet in the point R. The two circumcircles of ABCP and ACDR intersect at the point T. Prove that the points P, T, and R are collinear.

Solution:

Given:

ABCD is a cyclic quadrilateral. The two sides AB and DC are produced to meet in the point P and other two sides AD and BC are produced to meet in the point R. The two circumcircles of ABCP and ACDR intersect at the point T.

ABCD is a cyclic quadrilateral. The two sides AB and DC are produced to meet in the point P and other two sides AD and BC are produced to meet in the point R.

The two circumcircles of ΔBCP and ΔCDR intersect at point T.

Proof: Two angles in circle produced by the arc CP are ∠CBP and ∠CTP,

∴ ∠CBP = ∠CTP…..(1)

Again, ∠CDR and ∠CTR are two angles in circle produced by the chord CR.

∴ ∠CDR = ∠CTR ….(2)

Now, ABCD is a cyclic quadrilateral,

∴ ∠ABC + ∠ADC = 180°

or, 180°- ∠CBP + 180°- ∠CDR = 180° [∠ABC + ZCBP = 180° and ∠ADC + ∠CDR =. 180°]

or, ∠CTP + ∠CTR = 180° [from (1) and (2)]

∴ ∠PTR = 1 straight angle

Hence the three points P, T, R are collinear (Proved).

“WBBSE Class 10 Maths solved problems on cyclic quadrilaterals”

Example 12. O is the orthocentre of the AABC. Prove that O is also the incentre of its pedal triangle.

Solution:

Given:

O is the orthocentre of the AABC.

Let AD, BE and CF be three perpendiculars drawn from the vertices A, B and C of the ΔABC to their opposite sides BC, CA and AB respectively which intersect each other at O.

Then O is the orthocentre of the ΔABC. Let us join D, E; E, F and F, D.

Then ΔDEF is the pedal triangle of ΔABC.

To prove: O is the incentre of ΔDEF.

Proof: The four points A, C, D, and F are concyclic,

[since the two angles ∠DCF and ∠DAF on the same side of FD are equal.

∴ ∠DCF = 90° – ∠ABD = ∠BAD = ∠FAD]

Now, the two angles in circle produced by the arc AF are ∠ADF and ∠ACF,

∴ ∠ADF = ∠ACF

= 90° – ∠EAF [∠AFC = 90°, ∴ ∠ACF + ∠CAF = 90°]

= ∠ABE [∠AEB = 90°, ∴ ∠ABE + ∠EAB = 90°]

= ∠ADE [A, B, D, E are concyclic and ∠ABE and ∠ADE are two angles in circle produced by arc AE.]

i.e., ∠ODF = ∠ODE ∴ OD is the bisector of ∠EDF.

Similarly, it can be proved that OE and OF are the bisectors of ∠DEF and ∠DFE.

So, O lies on the bisectors of the angles of ADEF, i.e., O is the point of intersection of the bisectors of the angles of the ΔDEF.

Hence O is the incentre of ADEF. (Proved)

Example 13. ABCD is a cyclic quadrilateral such that AC bisects ∠BAD. AD is produced to E in such a way that DE = AB. Prove that CE = CA.

Solution:

Given:

ABCD is a cyclic quadrilateral such that AC bisects ∠BAD. AD is produced to E in such a way that DE = AB.

Given: ABCD is a cyclic quadrilateral. AC bisects ∠BAD. AD is produced to E in such a way that DE = AB.

To prove: CE = CA

Proof: ABCD is a cyclic quadrilateral.

∴ ∠ABC + ∠ADC = 180° . . . (1)

Again, ∠ADC + ∠CDE = 1 straight angle = 180° ……. (2)

From (1) and (2) we get, ∠ABC + ∠ADC – ∠ADC + ∠CDE

⇒ ∠ABC = ∠CDE……(3)

Now, in ΔABC and ΔCDE, DE = AB (given),

BC = CD [the angle in circle ∠BAC produced by the chord BC and the angle in circle ∠CAD produced by the chord CD are equal, ∴BC = CD.]

and included ∠ABC = included ∠CDE [from (3)]

∴ ΔABC ≅ ΔCDE [by the S-A-S condition of congruency]

∴ CA = CE [similar sides of congruent triangles]

Hence CA = CE. (Proved)

Example 14. In two circles, one circle passes through the centre O of the other circle and they intersect each other at the points A and B. A straight line passing through A intersect the circle passing through O at the point P and the circle with centre at O at the point R. By joining P, B and R. B prove that PR = PB.

Solution:

Given:

In two circles, one circle passes through the centre O of the other circle and they intersect each other at the points A and B. A straight line passing through A intersect the circle passing through O at the point P and the circle with centre at O at the point R.

Given: Let the circle with centre at C passes through the centre of other circle with centre at O.

The two circles intersect each other at the points A and B. A straight line PAR intersects the circle passing through O at the point P and also intersect the circle with centre at O at the point R.

The points P, B and B, R are joined.

To prove: PR = PB.

Construction: Let us join O, A; O, B and O, R

Proof: In the circle with centre at O, OB = OR [radii of same circle]

∴ ∠OBR = ∠ORB…… (1)

Again, PAOB is a cyclic quadrilateral,

∴ ∠PAO + ∠PBO = 180°  ……… (2)

Now, ∠PAO + ∠OAR = 1 straight angle =180°….(3)

Then from (2) and (3) we get, ∠PAO + ∠PBO = ∠PAO + ∠OAR

or, ∠PBO = ∠OAR or, ∠PBO = ∠ORA [OA = OR, ∠OAR = ∠ORA]……(4)

Now, ∠PBR = ∠PBO + ∠OBR  = ∠ORA + ∠ORB [from (4) and (1)] = ∠PRB.

∴ in ΔPBR , ∠PBR = ∠PRB, PR = PB. (Proved)

Example 15. Prove that cyclic parallelogram must be a rectangle.

Solution:

Let ABCD be a parallelogram inscribed in the circle with centre at O.

To prove: ABCD is a rectangle.

Proof: ABCD is a cyclic quadrilateral,

∴ ∠ABC + ∠ADC = 180° …….(1)

Again, ABCD is a parallelogram.

∴ ∠ABC = ∠ADC [opposite angles of a parallelogram are equal.]

∴  from (1) we get, ∠ADC + ∠ADC = 180° or, 2 ∠ADC = 180° or, ∠ADC = 90°

Similarly, it can be proved that ∠ABC = ∠BCD = ∠CDA = ∠DAB = 90°

Also, since ABCD is a parallelogram,

∴ AB = DC and AD = BC.

Hence ABCD is a rectangle. (Proved)

“Cyclic quadrilateral angle relationships for Class 10 students”

Example 16. Prove that any four vertices of a regular pentagon are concyclic.

Solution:

Let ABCDE he a regular pentagon.

∴ AB = BC = CD = DE = EA.

To prove: Any four let A, B, C, E points are concyclic.

Construction: Let us join A, C and B, E.

Proof: In triangles ABC and ABE, BC = AE, AB is common to both and ∠ABC = ∠BAE [angles of a regular pentagon are equal.]

∴ ΔABC = ΔABE [by the S-A-S condition of congruency]

∴ ∠ACB = ∠AEB,

i.e. the angles produced at two points C and E on the same side of AB are equal.

A, B, C, E are concyclic.

Similarly, by selecting any other four points, it can be proved that the selected four points are concyclic.

Hence any four vertices of a regular pentagon are concyclic. (Proved)

Example 17. ABCD is a cyclic quadrilateral. The side BC of it is extended to E. Prove that the two bisectors of ∠BAD and ∠DCE meet on the circumference of the circle.

Solution:

Given:

ABCD is a cyclic quadrilateral. The side BC of it is extended to E.

Let ABCD is a cyclic quadrilateral in the circle with centre at O. The side BC of it is extended to E.

To prove: The two bisectors of ∠BAD and ∠DCE meet on the circumference of the circle.

Construction: Let us construct the bisector AP of ∠BAD, in which A, intersect the circle at P.

Let us join P, C and PC is extended to Q.

Proof: ∠BCP = ∠QCE [opposite angles] ……….. (1)

Now, ADCP is a cyclic quadrilateral.

∴ ∠PAD + ∠PCD = 180°……(2)

But ∠PCD + ∠DCQ =180°…… (3)

∴ from (2) and (3) we get, ∠PAD + ∠PCD = ∠PCD + ∠DCQ or, ∠PAD = ∠DCQ

or, ∠PAB = ∠DCQ…….. (4) [AP is the bisector of ∠BAD]

or, ∠PCB = ∠DCQ [both ∠PAB and ∠PCB are angles in circle produced by the arc BP.]

or, ∠QCE = ∠DCQ [from (1)] or, ∠ECQ = ∠DCQ

∴ CQ is the bisector of ∠DCE.

i.e…., PQ is the bisector of ∠DCE and intersects the circle at P.

Hence the two bisectors of ∠BAD and ∠DCE meet on the circumference of the circle. (Proved)

Example 18. AB is a diameter of a circle. PQ is such a chord of the circle that it is neither a diameter of the circle nor an interceptor of AB. By joining the points A, P and B, Q it is found that ΔBQP is a quadrilateral of which ∠BAP = ∠ABQ. Prove that ΔBQP is a cyclic trapezium.

Solution:

Given:

AB is a diameter of a circle. PQ is such a chord of the circle that it is neither a diameter of the circle nor an interceptor of AB. By joining the points A, P and B, Q it is found that ΔBQP is a quadrilateral of which ∠BAP = ∠ABQ.

Let AB be the diameter of the circle with centre at O and PQ is such a chord of the circle that is neither a diameter of the circle nor an interceptor of AB.

ABQP is a quadrilateral of which ∠ZBAP = ∠ABQ.

To prove: ABQP is a cyclic trapezium.

Proof: ABQP is a cyclic quadrilateral.

∴ ∠BAP + ∠BQP = 180°……..(1)

and ∠ABQ + ∠APQ =180°…….(2)

From (1) and (2) we get, ∠BAP + ∠BQP = ∠ABQ + ∠APQ

or, ∠BAP + ∠BQP = ∠BAP + ∠APQ [∠ABQ = ∠BAP]

or, ∠BQP = ∠APQ

from (1) we get, ∠BAP + ∠APQ = 180° [∠BQP = ∠APQ]

Thus, the sum of two adjacent angles on the same side of the transversal AP of two line segments AB and PQ is 180°.

∴ AB and PQ are parallel to each other, i.e., two opposite sides AB and PQ of the cyclic quadrilateral ABQP are parallel to each other.

Hence ABQP is a cyclic trapezium. (Proved)

Example 19. ABCD is a cyclic quadrilateral; The bisectors of ∠A and ∠C intersect the circle at points E and F respectively. Prove that EF is the diameter of the circle.

Solution:

Given:

ABCD is a cyclic quadrilateral; The bisectors of ∠A and ∠C intersect the circle at points E and F respectively.

Let ABCD is a cyclic quadrilateral in the circle with centre at O.

The bisector AE of the ∠A intersects the circle at E and the bisector of the ∠C intersects the circle at F. Let us join E, F.

To prove: EF is a diameter of the circle.

Construction: Let us join C, E.

Proof: ABCD is a cyclic quadrilateral.

∴ ∠BAD + ∠BCD = 180°

or, \(\frac{1}{2}\) ∠BAD + \(\frac{1}{2}\) ∠BCD = 90° [dividing by 2]

or, ∠DAE + ∠DCF = 90° [BE and CF are the bisectors of ∠A and ∠C respectively]

But two angles in circle produced by the arc DE are ∠DAE and ∠DCE.

∴ ∠DAE = ∠DCE.

∴ from (1) we get, ∠DCE + ∠DCF = 90° or, ∠ECF = 90°

∴ ∠ECF is a semicircular angle.

Hence EF is the diameter of the circle. [Proved]

Example 20. ΔABC is an acute angled triangle inscribed in a circle. AD is a diameter of the circle. Two perpendiculars BE and CF are drawn from B and C to AC and AB respectively, which intersect each other at the point G. Prove that BDCG is a parallelogram.

Solution:

Given:

ΔABC is an acute angled triangle inscribed in a circle. AD is a diameter of the circle. Two perpendiculars BE and CF are drawn from B and C to AC and AB respectively, which intersect each other at the point G.

Let ΔABC is an acute angled triangle, inscribed in a circle with centre at O.

Two perpendiculars BE and CF are drawn from B and C to the sides AC and AB which intersect each other at the point G.

AD is a diameter of the circle.

To prove: BDCG is a parallelogram.

Proof: AD is a diameter, ∴ ∠ABD = 90° [semicircular angle]

Again two angles in circle produced by the chord BD are ∠BCD and ∠BAD.

∴ ∠BCD = ∠BAD [angles in the same segment of circle are equal.]

= 90° – ∠BDA [∠ABD = 90°, ∠BAD + ∠BDA = 90°]

= 90° – ∠ACB [∠BDA and ∠ACB are angles in circle produced by the arc AB.]

= ∠CBE [∠BEC = 90°, ∴ ∠CBE + ∠BCE = 90°] = ∠CBG .

i.e., ∠BCD = ∠CBG.

But these are alternate angles, BG || DC.

Similarly, it can be proved that BD || GC. So, two opposite sides of the quadrilateral BDCG are parallel.

Hence BDCG is a parallelogram. (Proved)