WBBSE Solutions For Class 9 Maths Solid Geometry Chapter 1 Properties Of Parallelogram

WBBSE Solutions For Class 9 Maths Solid Geometry Chapter 1 Properties Of Parallelogram

WBBSE Class 9 Properties of Parallelograms Overview

Chapter 1 Properties Of Parallelogram Introduction

1. In the previous chapters, you have learned what is geometry.
2. Who are the pioneers of geometry?
3. What are points and straight lines?
4. What are straight lines, line segments, and rays?
5. What are intersecting straight lines, and parallel straight lines?
6. What are concurrent St. lines, what are transversals, etc.?
7. What are angles and their different types and properties?
8. What are triangles and their various kinds and properties?
9. What are polygons and their various kinds and properties?
10. The concepts of congruence of the triangle?

Read and Learn More WBBSE Solutions For Class 9 Maths

You have already seen that if the opposite sides of a quadrilateral be parallel to each other, then the quadrilateral is a parallelogram.

Basic Properties of Parallelograms

The main properties of a parallelogram are :

1. Each diagonal bisects the quadrilateral into two equal triangles.
2. The opposite sides of a parallelogram are equal.
3. The opposite angles of a parallelogram are equal.
4. The diagonals of a parallelogram bisect each other.

We shall now prove the theories regarding these properties one by one.

Theorem 1. In Any parallelogram.

1. Every diagonal of a parallelogram bisects it into two congruent triangles.

2. The lengths of the opposite sides of a parallelogram are equal.

3. The measures of the opposite angles are equal.

Given:

Let ABCD be a parallelogram, in which AB || DC and AD || BC. 

The diagonal AC has bisected the parallelogram ABCD into two triangles ΔABC and ΔACD.

To prove: 

We have to prove that

1. ΔABC=ΔCDA

2. AB = DC; BC= AD

3. ∠ABC = ∠ADC; ∠BAD = ∠BCD

Prove:

1. In A’s ABC and CDA,

∠BAC alternate ∠ACD……………..(1) [ AB || DC and AC is their transversal] 

AC is common to both the triangles and ∠ACB = alternate CAD..…………..(2)

[For some cause]

∴ ΔABC ≅ ΔCDA        [ by the ASA congruence]      (Proved)

2. From (1) we get, ΔABC ≅ ΔCDA

∴ AB = DC (∵ similar sides)

and BC = AD (∵ similar sides) (Proved) 

3. From (1) we get, ΔABC ≅ ΔCDA.

∴ ∠ABC = ∠ADC ( similar angles) 

Again, ∠BAC+ ∠CAD = ∠ACD + ∠ACB [from (1) and (2)]

Or, ∠BAD =∠BCD.

∴ ∠ABC = ∠ADC

and ∠BAD = ∠BCD.      (Proved)

Theorems Related to Parallelograms

Theorem 2. The Diagonals Of A Parallelogram Bisect Each Other

Given: ABCD is a parallelogram in which AB || DC and AD || BC. 

The two diagonals AC and BD intersect each other at point O.

To prove: We have to prove that, AO = OC and BO = OD.

Proof: In triangles, ΔAOD andΔBOC, 

∠OAD = ∠OCB 

[ ∵ AD | BC and AC is their transversal, ∴ ∠CAD = alternate ∠ACB or, ∠OAD = ∠OCB] 

AD = BC  [∵ Opposite sides of the parallelogram ABCD]

and ∠ODA = ∠OBC

[∵ AD || BC and BD is their transversal, ∴ ∠BDA = alternate ∠DBC or, ∠ODA = ∠OBC]

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

∴ AO = OC (similar sides)

and BO = OD (similar sides).     (Proved)

Theorem 3. If The Opposite Sides Of A Quadrilateral Are Equal In Length, The Quadrilateral Is A Parallelogram

Diagonals of a Parallelogram

Given: In the quadrilateral ABCD, AB = DC and AD = BC.

To prove: We have to prove that ABCD is a parallelogram. Construction Join the diagonal BD.

Proof: In triangles AABD and ABCD.

AB = DC and AD BC (Given) and BD is common to both.

∴ ΔABD ≅  ΔBCD [ by the condition of S-S-S congruence ]

∴ ∠ADB = ∠CBD [∵ similar angles]

But since BD intersects both AD and BC, these are alternate angles and are equal.

∴ AD || BC.

Again, ∠ABD = ∠CDB [∵ similar angles]

But, since BD intersects both AB and DC, these are alternate angles and are equal in magnitude. 

∴ AB || DC.

Thus, in the quadrilateral ΔBCD, AD || BC and AB || DC.

Hence, ABCD is a parallelogram.    (Proved)

Theorem 4. If The Opposite Angles Of A Quadrilateral Are Equal In Magnitude, Then The Quadrilateral Is A Parallelogram

Given: In the quadrilateral ABCD, ∠BAD = ∠BCD and ∠ABC = ∠ADC.

To prove: We have to prove that ABCD is a parallelogram.

Proof :

∠BAD = ∠BCD……………………….(1) (Given)

∠ABC = ∠ADC……………………..(2) (Given)

Now, ∠ABC + ∠BCD + ∠CDA + ∠DAB = 360°

[∵ the sum of the four angles of any quadrilateral is 360°]

or, ∠ABC + ∠BAD+ ∠ABC + ∠BAD = 360° [From (1) and (2)]

or, 2(∠ABC + ∠BAD) = 360° or, ∠ABC + ∠BAD = 180° [Dividing by 2]

∴ AD || BC

[∵ if the sum of the internal angles of the two angles on the same side of the transversal of two straight lines is 180°, then the straight lines are parallel. ]

Similarly, it can be proved that AB || DC.

Thus, in the quadrilateral ABCD, AB || DC and AD || BC, i.e., the opposite sides are parallel. 

Hence, ABCD is a parallelogram. (Proved)

Theorem 5. If Each Pair Of Opposite Sides Of Any Quadrilateral Is Parallel And Equal In Length, Then The Quadrilateral Is A Parallelogram

Given: In the quadrilateral ABCD, AB = DC and AB || DC.

To prove: We have to prove that ABCD is a parallelogram.

Construction: Let us join the diagonal AC.

Proof: In triangles, ΔABC and ΔCDA, AB = DC (Given)

∠BAC = alternate ∠ACD [∵ AB || DC and AC are their transversal ]

and the side AC is common to both.

∴ ΔABC ≅ ΔCDA [by the condition of S-A-S congruence]

∴ ∠ACB =∠DAC [∵ similar angles]

But, since AC intersects both AD and BC, these are alternate angles, which are equal in magnitude. 

∴ AD || BC.

Thus, in the quadrilateral ABCD, AB || DC and AD || BC.

∴ ABCD is a parallelogram. (Proved)

Theorem 6. If The Diagonals Of A Quadrilateral Bisect Each Other Equally, Then The Quadrilateral Is A Parallelogram

Given: In quadrilateral ABCD, the diagonals AC, and BD intersects each other equally at point O,

i.e., AO = OC and BO = OD.

To prove: We have to prove that ABCD is a parallelogram.

Proof: In triangles, ΔAOB and ΔCOD, AO = OC (Given),

BO = OD (Given)

and ∠AOB = ∠COD [opposite angles]

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

∴ AB = CD similar sides and

∠OAB = ∠OCD [ ∵ similar angles]

But, AC intersects both AB and DC for which these two are alternate angles and are equal.

∴ AB || DC.

Thus, two opposite sides AB and DC of the quadrilateral ABCD are equal and parallel

∴ ABCD is a parallelogram. (Proved)

We shall now discuss the application of the above theorems in various real problems in the following examples.

Chapter 1 Properties Of Parallelogram Selecte The Correct Answer(MCQ)

Question 1. If BAD = 75″ and CBD = 60″ in the parallelogram ABCD, then the measurement of <BDC is-

1. 60°
2. 75°
3. 45°
4. 50°

Solution: ∠BCD = ∠BAD = 75° [∵opposite angles of parallelograms are equal ]

Now, ∠BDC + ∠BCD + ∠CBD = 180°

or, ∠BDC+75° + 60° + 180° 

or, ∠BDC = 45°

∴ <BDC = 45°.

Question 2. In which one of the following geometric figures, the length of two diagonals are equal?

1. Parallelogram
2. Rhombus
3. Trapezium 
4. Rectangle

Solution: The lengths of two diagonals of a rectangle are equal.

∴ 4. Rectangle is correct.

 

Question 3. In the parallelogram ABCD, M is the mid-point of the diagonal BD. BM bisects equally ABC, then the measurement of ZAMB =

1. 45°
2. 60°
3. 90°
4. 75°

Solution: Given that ABCD is a parallelogram,

∴ AD || BC and BD are their transversals.

∴ ∠ADB alternate

or, ∠ADB = ∠ABD [ BM, is the bisector of ∠ABC]

∴ AB = AD in ΔABD i.e., ΔABD is an isoscele triangle. 

Again, M is the mid-point of BD.

∴ AM ⊥ BD. 

∴ ∠AMB = 90°.

Question 4. If ∠ACB = 40° in rhombus ABCD, then ∠ADB =

1. 50°
2. 110°
3. 90°
4. 120°

Solution:

Let the diagonals AC and BD of rhombus ABCD intersects each other at point O.

∴ ∠BOC = 90°

[∵ The diagonals of any rhombus intersect each other equally at right angles.]

Now, ∠BOC+∠OCB + ∠OBC= 180°

or, 90° + 40°+ ∠OBC= 180° 

or, ∠OBC = 50°.

But, ∠OBC = DBC= alternate ∠ADB. 

∴ ∠ADB = 50°.

Applications of Parallelogram Properties in Geometry

Chapter 1 Properties Of Parallelogram Short Answer Type Questions

Question 1.  If in the parallelogram ABCD ∠A: ∠B = 3:2, then find the angles of the parallelogram ABCD.

Solution:

∠A and B in the parallelogram ABCD are two adjacent angles.

∴ ∠A+ ∠B = 180°……………………..(1) 

[∵ the sum of the two internal angles of the transversal of any two parallel sides of a parallelogram on one side is 180°]

Again, ∠A: ∠B = 3:2 [ Given ]

\(\frac{\angle \mathrm{A}}{\angle \mathrm{B}}=\frac{3}{2} \text { or, } 2 \angle \mathrm{A}=3 \angle \mathrm{B} \quad \text { or, } \angle \mathrm{A}=\frac{3 \angle \mathrm{B}}{2}\)………………….(2)

Putting ∠A= \(\frac{3 \angle B}{2}\) in (1) we get, \(\frac{3 \angle \mathrm{B}}{2}+\angle \mathrm{B}=180^{\circ}\)

∴ \(\frac{5 \angle \mathrm{B}}{2}=180^{\circ} \quad \text { or, } \angle \mathrm{B}=\frac{180^{\circ} \times 2}{5} \text { or, } \angle \mathrm{B}=72^{\circ}\)

Also putting ∠B = 72° in (2) we get, ∠A= \(\angle \mathrm{A}=\frac{3 \times 72^{\circ}}{2}=108^{\circ}\)

Now, ∠C = ∠A (opposite angles of a parallelogram are equal) = 108°.

∠D = ∠B (for same cause) = 72°

∴ the angles of ABCD are∠A = 108°, ∠B = 72°, ZC = -108°, ∠D = 72°.

Question 2. The bisectors of ∠A and B meet at point E on the side CD of the parallelogram ABCD. If the length of side BC is 2 cm, find the length of side AB.

Solution:

∴ ABCD is a parallelogram, 

∴ ∠A + ∠B = 180° [∵ ZA and ZB are two adjacent angles.]

∴ \(\frac{1}{2} \angle A+\frac{1}{2} \angle B\) = 90° or, ∠EAB + ∠EBA = 90°

∴∠AEB 180° – (∠EAB + ∠EBA) = 180°-90° = 90°

∴ AE and BE are half of the diagonals of such a rhombus, one of which side is AB and half of the adjacent other sides BC= 2 cm.

∴ that side of the rhombus = 2BC= 2 x 2 cm = 4 cm.

AB = BC = 2BC= 4 cm,, 

∴ AB = 4 cm.

The length of side AB = 4 cm.

Alternative Method: Let us construct AD || FE. 

Let EF intersect AB at F.

Now, FB || EC, and BE is their transversal.

∵ ∠F || BE alternate ∠CEB.

or, ∠EBC = ∠CEB [ ∵ BE is the bisector of ∠B   ∴ ∠FBE = ∠EBC.]

∴ CE = BC=2 cm [∵ opposite sides of equal angles are equal. ] 

Similarly, it can be proved that DE = 2 cm.

∴ CD DE+ CE=2 cm + 2 cm = 4 cm. 

∴ AB=4 cm [ AB and CD are opposite sides of the parallelogram ABCD and they are equal.]

Examples of Parallelogram Problems with Solutions

Question 3. If the equilateral triangle ΔAOB is into the square ABCD. Find ∠COD.

Solution:

ΔAOB is equilateral, 

∴ AB = BO = OA and ∠AOB = ∠OAB = ∠OBA = 60°. 

Again, BC = AB = BO, 

∴ ∠BOC = ∠BCO.

Now, in triangle BOC, ∠BOC + ∠BCO + ∠OBC= 180°.

or, ∠BOC+ ∠BOC+ (90° – ∠OBA) = 180° [∵ ∠BCO =  ∠BOC]

or, 2∠BOC + 90° – 60° = 180° [∵ ∠OBA = 60°]

or, 2∠BOC = 150° 

or, ∠BOC = 75°

Similarly, ∠AOD = 75°

∴ ∠COD 360° – (∠AOD + ∠AOB + ∠BOC)

= 360° – (75° + 60° +75°)

= 360° – 210°

= 150°.

∠COD = 150°.

Question 4. M is a point on the side AD of the square ABCD such that CMD= 30°. If the diagonal BD intersects CM at a point P, then find ∠DPC.

Solution:

∠DPC = internally opposite [∠PDM + ∠PMD]……………..(1) 

Now, ∠ADB = ∠ABD [ ∵ AB = AD]

Again, BD bisects ∠ADC.

∠BDA = \(\frac{90°}{2}\)

or, ∠PDM = 45°

∴ we get from (1), ∠DPC = 45° + 30° 75°, [ ∵ ∠PMD = 30° ]

Question 5. The length of AB of the rhombus ABCD is 4 cm and if ∠BCD = 60″, then find the length of the diagonal BD.

Solution:

In ΔBCD, BC= CD [ ∵ ABCD is a rhombus]

∴ ∠BDC= DBC [ ∵ opposite angles of equal sides]

∴ In ΔBCD, BDC + 2DBC + ∠BCD = 180°

or, ∠BDC + ∠BDC + 60° 180° 

or, 2∠BDC = 120°

or, BDC = 60° = <DBC          [∵ ∠BDC = <DBC.]

∴ ABCD is equilateral,

∴ BC= CD = DB.

Again, BC = CD = AB = 4 cm (Given). 

∴ BD = 4 cm.

The length of the diagonal BD = 4 cm.

Question 6. In the parallelogram, ABCD, AP, and DP are the bisectors of the BAD and ∠ADC respectively. Find ∠APD.

Solution:

∠A and ∠D, are the two adjacent angles of

the parallelogram ABCD.

∴ ∠A+ ∠D = 180°

Or, \(\frac{1}{2} \angle \mathrm{A}+\frac{1}{2} \angle \mathrm{D}=90^{\circ}\)

or, ∠PAD+ ∠PDA = 90°

∴ ∠APD = 180° – (∠PAD+ ∠PDA)

= 180°-90°

= 90°.

∠APD = 90°.

Chapter 1 Properties Of Parallelogram Long Answer Type Questions

Question 1. Prove that if the lengths of two diagonals of any parallelogram be equal and if the diagonals intersect each other orthogonally, then the parallelogram is a square.

Given: The diagonals PR and QS of the parallelogram are such that PR = QS, PR ⊥ QS, and O is the point of intersection of PR and QS.

To prove: We have to prove that PQRS is a square.

Proof: In A’s ΔPOQ and ΔQOR, OP = OR [ diagonals of any parallelogram bisects each other. ]

OQ is common to both triangles and

∠POQ = ∠QOR [each are 90°.]

∴ ΔPOQ=ΔQOR [ by the condition of S-A-S congruence].

PQ = QR [ similar sides]

∴ PQ = QR = RS = SP. [ PQ= SR and QR = PS]……………….(1) 

Again, in triangles ΔPQS and ΔPQR, PS = QR (Proved) QS = PR [Given], and PQ are common to both the triangles.

ΔPQS ≅ ΔPQR [by the condition of S-S-S congruence]

∴ ∠QPS = ∠PQR [ similar angles]

Now, ∠QPS+∠PQR = 180° 

[ PS || QR and PQ are transversal so these are the internal angles on the same side of PQ.] 

or, ∠QPS+∠QPS=180°

or, 2∠QPS = 180° 

or, ∠QPS = 90°

∴ PQR = ∠QRS = ∠RSP = ∠SPQ=90° ……..(2)

From (1) and (2) we get, PQRS as a square. (Proved)

Visual Representation of Parallelogram Properties

Question 2. Prove that if the lengths of the two diagonals of a parallelogram be equal, then the parallelogram is a rectangle.

Given: Let the diagonals AC and BD of the parallelogram ABCD be such that AC = BD.

To prove: We have to prove that ABCD is a rectangle.

Proof: In triangles ΔACD and ΔBCD.

AD = BC [ opposite sides of ABCD]

DC is common to both the triangles and AC = BD (Given) 

∴ ΔACD=ΔBCD [ by the condition of S-S-S congruence] 

∴ ∠ADC=∠BCD [ similar angles of congruent triangles] 

But, these are two adjacent angles of the parallelogram ABCD.

∴ ∠ADC+∠BCD= 180° 

or, ∠ADC+∠ADC= 180°

or, 2∠ADC= 180° 

or, ∠ADC=90° each and every angle of ABCD is 90°.

∴ ABCD is a rectangle. (Proved)

Question 3. Prove that if the two diagonals of any parallelogram intersect each other orthogonally, then the parallelogram is a rhombus.

Given: Let the two diagonals AC and BD of the parallelogram ABCD intersect each other orthogonally at the point O,

i.e., AO = OC, BO = OD and ∠AOB = ∠BOC=∠COD=∠DOA = 90°.

To prove: ABCD is a rhombus.

Proof: In triangles, ΔAOB and ΔBOC, AO = OC (Given),

OB is common to both the triangles and ∠AOB = ∠BOC [∵ each are 90° ], 

ΔAOB ≅ ΔBOC [by the condition of S-A-S congruence],. 

∴ AB = BC [ ∵ similar sides] 

∴ in the parallelogram ABCD, AB = BC= CD=DA [∵ AB = DC, AD = BC]

∴ ABCD is a rhombus. (Proved)

Question 4. Prove that the two angles attached to the parallel sides of any isosceles trapezium are equal to each other.

Given: Let ABCD be an isosceles trapezium of which AB || DC and AD = BC.

To prove: ∠ADC= <BCD and DAB = <CBA.

Construction: Let us draw perpendiculars AE and BF from A and B respectively to the side DC of ABCD, which intersect DC at E and F respectively.

Proof: In the right-angled triangles. ΔADE and ΔBCF, ∠AED = ∠BFC [∵ each are right angles] 

hypotenuse AD = hypotenuse BC (Given)

and AE = BF [ the two heights of ABCD are equal]

∴ AADE ≅ ABCF [∵  by the condition of R-H-S congruence]

∴ ∠ADE = ∠BCF [ ∵  similar angles of congruent triangles ]

or, ∠ADC= ∠BCD.

Again, ∠DAE=∠CBF [∵ similar angles of congruent triangles]

or, ∠DAE+∠EAB = CBF + FBA [∵ ∠EAB = ∠FBA = 90°]

or, ∠DAB = ∠CBA.

∴ ∠ADC= ∠BCD and 2DAB = <CBA. (Proved)

Question 5. The diagonals AC and BD of the parallelogram ABCD intersect each other at O. Any straight line passing through O intesects the sides AB and DC at the points P and Q respectively. Prove that OP=OQ.

Given: The diagonals AC and BD of the parallelogram ABCD intersect each other at O.

The straight line PQ passing through O, intersect AB at P and DC at Q.

To prove: OP=OQ.

Proof: In triangles, ΔAOP and ΔCOQ, ∠APO = <CQO [∵ alternate angles]

<PAO = ∠QCO [∵ alternate angles]

and AO = OC[ the diagonals of any parallelogram bisect each other.] 

∴ ΔAOP ≅ ΔCOQ [by the conditions of A-A-S congruence] 

∴ OP = OQ [∵ similar sides of congruence triangles] (Proved)

Question 6. P is any point on BC of the square ABCD. The perpendicular, drawn from B to AP intersects DC at Q. Prove that AP = BQ. 

Given: P is any point on BC of the square ABCD. 

The perpendicular BM, drawn from B to AP intersects DC at Q.

To prove AP = BQ.

Proof: BM ⊥ AP, 

∴ BMP = 90°

Now, in ΔABP, ∠ABP +∠APB + ∠PAB = 180°

or, 90° + ∠APB + ∠PAB = 180°

or, ∠APB + ∠PAB = 90°………….(1)

Again, in, A BMP, ∠BMP + ∠BPM + ∠PBM = 180° 

or, 90°+ ∠BPM + ∠PBM = 180°

or, ∠APB +∠PBM=90° [ ∠BPM = ∠APB]…………..(2)

Then, from (1) and (2) we get, ∠APB+ ∠PAB = ∠APB + ∠PBM.

or, ∠PAB = ∠PBM……………………..(3)

Now, in triangles, ΔABP and Δ BCQ, PAB = <CBQ [∵ ∠PBM=CBQ and from (3)] 

Also, ∠ABP = ∠BCQ [ ∵ each are 90° ]

and AB = BC [∵each are the sides of the square ABCD]

∴ ΔABP ≅ ΔBCQ [ by the condition of A-A-S congruence]

∴ AP = BQ (Proved) [ ∵ similar sides of congruent triangles ]

Question 7. The two medians BP and CQ of ΔABC are extended to R and S respectively in such a way that BP = PR and CQ = QS. Prove that S, A, R are collinear, i.e., lie on the same straight line. Given: The medians BP and CQ of ΔABC are extended to R and S respectively such that BP = PR and CQ = QS.

To prove: S, A, R are collinear.

Construction: Let us join S, A; A, R; B, S, and C, R.

Proof: The diagonals AC and BR of the quadrilateral ABCR bi-sect each other at P, since AP = CP and BP = PR,

∴ ABCR is a parallelogram.

∴ AR || BC……..(1) [∵ opposite sides ]

Again, the diagonals AB and CS of the quadrilateral ACBS bisect each other, since AQ=BQ and CQ = SQ

∴ SA || BC ………..(2) [∵ opposite sides ]

∴ ACBS is a parallelogram.

Then from (1) and (2) we get, SA || AR.

But the common end points SA and AR is A.

∴ S, A, R are collinear. (Proved)

Question 8. AC is a common diagonal to both the parallelograms ABCD and AECF. If the points B, E, D, F are not collinear, prove that BEDF is a parallelogram.

Given: AC is a common diagonal of both the parallelograms ABCD and AECF and the points B, E, D, F are not collinear.

To prove: BEDF is a parallelogram.

Proof: In triangles, ΔADE and ΔBCF.

AE = CF [ opposite sides of a parallelogram]

AD = BC [ for same reasons]

and ∠DAE = ∠BCF [∵ ∠DAE =<DAC + ZEAC = alternate BCA + alternate ∠FCA = ∠BCF]

∴ AADE = ABCF [∵ by the condition of S-A-S congruent]

⊂ DE FB………..(1) [∵ similar sides of congruent triangles]

Again, in triangles, AADF and ABCE, AD = BC [∵ opposite angles of a parallelogram] 

AF = CE [∵ for same reasons ] and ∠DAF = ∠BCE 

[… ∠DAC = alternate ∠BCA…………(1). ∠FAC alternate ∠ECA………..(2)  ∴ (1) – (2)=<DAC-∠FAC = ∠BCA-∠ECA, or, ∠DAF = <BCE] 

AADF ≅ ABCE [∵ by the condition of S-A-S congruence ]

∴ DF = BE.…………..(2) [∵ similar sides of congruent triangles]

Then we get, from (1) and (2) in the quadrilateral BEDF, ED = BF and BE = DF,

i.e., two pairs of opposite sides of the quadrilateral BEDF are equal to each other.

∴ BEDF is a parallelogram. (Proved)

Question 9. ABCD is a quadrilateral. ABCE and BADF are two constructed parallelograms. Prove that CD and EF bisects each other equally.

Given: ABCD is a quadrilateral. 

Two parallelograms ABCE and BADF are constructed.

To prove: CD and EF bisects each other equally.

Construction: Let us join C, D and E, F.

Proof : ABCE is a parallelogram

∴ AB || EC and ABEC………....(1)

Again, BADF is a parallelogram,

∴ AB || DF and AB = DF………………….(2)

Now, from (1) and (2) we get, EC || DF and EC = DF

∴ CEDF is a parallelogram [∵ two opposite sides are equal and parallel]

∴ two diagonals of CEDF bisects each other equally. (Proved)

Question 10. In the parallelogram ABCD, AB = 2AD. Prove that the bisectors of the angles <BAD and ∠ABC meet at the mid-point of DC at right-angles.

Given: In the parallelogram ABCD, AB = 2AD. 

The bisectors AP and BP of the angles ∠BAD and ∠ABC respectively, meet at a point P on DC.

To prove: 

1. P is the mid-point of DC. 

2. ∠APB = 90°.

Proof: 

1. AB || DC and AP is their transversal.

∴ ∠BAP = ∠DPA     [∵ alternate angle ]………. (1)

Again, ∠BAP = ∠DAP……………….(2) [ ∵ AP bisects <BAD]

Now, from (1) and (2) we get, ∠DPA =<DAP.

AD = DP [opposite sides of equal angles ] 

or, \(\frac{1}{2}\) AB = DP [∵ AB = 2AD] 

or, \(\frac{1}{2}\) DC=DP [∵ AB = DC]

or, DP =  \(\frac{1}{2}\) DC. 

∴ P is the mid-point of DC. (Proved)

2. Since ∠DAB and ∠CBA are two internal angles of the same side of the transversal AB of the parallel straight lines AD and BC, we get

<DAB+ <CBA = 180° 

or, \(\frac{1}{2}\) <DAB +  \(\frac{1}{2}\) <CBA = 90° 

or, ∠PAB+∠PBA = 90° [∵ AP and BP are the bisectors of ∠DAB and CBA respectively] 

Now, in triangle APB, ∠APB +∠PAB + <PBA = 180°.

or, ∠APB +90° = 180° [∵ ∠PAB+ ∠PBA = 90°]

∠APB = 90°. (Proved)

∴ ∠CBA-90°