NCERT Solutions for Exercise 8.1 Class 9 Maths Chapter 8 - Quadrilaterals

Access Quadrilaterals Class 9 Chapter 8 Exercise: 8.1

Q1 The angles of quadrilateral are in the ratio $3:5:9:13$ . Find all the angles of the quadrilateral.

Answer:

Given : The angles of a quadrilateral are in the ratio $3:5:9:13$ .
Let the angles of quadrilateral be $3x,5x,9x,13x$ .

Sum of all angles is 360.

Thus, $3x+5x+9x+13x={360}^{\circ }$

$\Rightarrow 30x={360}^{\circ }$

$\Rightarrow x={12}^{\circ }$

All four angles are : $3x=3\times 12={36}^{\circ }$

$5x=5\times 12={60}^{\circ }$

$9x=9\times 12={108}^{\circ }$

$13x=13\times 12={156}^{\circ }$

Q2 If the diagonals of a parallelogram are equal, then show that it is a rectangle.

Answer:

Given: ABCD is a parallelogram with AC=BD.

To prove: ABCD is a rectangle.

Proof: In $\mathrm{△}$ ABC and $\mathrm{△}$ BAD,

BC= AD (Opposite sides of parallelogram)

AC=BD (Given )

AB=AB (common)

$\mathrm{△}$ ABC $\cong$ $\mathrm{△}$ BAD (By SSS)

$\mathrm{\angle }ABC=\mathrm{\angle }BAD$ (CPCT)

and $\mathrm{\angle }ABC+\mathrm{\angle }BAD={180}^{\circ }$ (co - interior angles)

$2\mathrm{\angle }BAD={180}^{\circ }$

$\mathrm{\angle }BAD={90}^{\circ }$

Hence, it is a rectangle.

Q3 Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.

Answer:

Given: the diagonals of a quadrilateral bisect each other at right angles i.e.BO=OD.

To prove: ABCD is a rhombus.

Proof : In $\mathrm{△}$ AOB and $\mathrm{△}$ AOD,

$\mathrm{\angle }AOB=\mathrm{\angle }AOD$ (Each ${90}^{\circ }$ )

BO=OD (Given )

AO=AO (common)

$\mathrm{△}$ AOB $\cong$ $\mathrm{△}$ AOD (By SAS)

AB=AD (CPCT)

Similarly, AB=BC and BC=CD

Hence, it is a rhombus.

Q4 Show that the diagonals of a square are equal and bisect each other at right angles.

Answer:

Given : ABCD is a square i.e. AB=BC=CD=DA.

To prove : the diagonals of a square are equal and bisect each other at right angles i.e. AC=BD,AO=CO,BO=DO and $\mathrm{\angle }COD={90}^{\circ }$

Proof : In $\mathrm{△}$ BAD and $\mathrm{△}$ ABC,

$\mathrm{\angle }BAD=\mathrm{\angle }ABC$ (Each ${90}^{\circ }$ )

AD=BC (Given )

AB=AB (common)

$\mathrm{△}$ BAD $\cong$ $\mathrm{△}$ ABC (By SAS)

BD=AC (CPCT)

In $\mathrm{△}$ AOB and $\mathrm{△}$ COD,

$\mathrm{\angle }$ OAB= $\mathrm{\angle }$ OCD (Alternate angles)

AB=CD (Given )

$\mathrm{\angle }$ OBA= $\mathrm{\angle }$ ODC (Alternate angles)

$\mathrm{△}$ AOB $\cong$ $\mathrm{△}$ COD (By AAS)

AO=OC ,BO=OD (CPCT)

In $\mathrm{△}$ AOB and $\mathrm{△}$ AOD,

OB=OD (proved above)

AB=AD (Given )

OA=OA (COMMON)

$\mathrm{△}$ AOB $\cong$ $\mathrm{△}$ AOD (By SSS)

$\mathrm{\angle }$ AOB= $\mathrm{\angle }$ AOD (CPCT)

$\mathrm{\angle }$ AOB+ $\mathrm{\angle }$ AOD = ${180}^{\circ }$

2. $\mathrm{\angle }$ AOB = ${180}^{\circ }$

$\mathrm{\angle }$ AOB = ${90}^{\circ }$

Hence, the diagonals of a square are equal and bisect each other at right angles.

Q5 Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.

Answer:

Given : ABCD is a quadrilateral with AC=BD,AO=CO,BO=DO, $\mathrm{\angle }$ COD = ${90}^{\circ }$

To prove: ABCD is a square.

Proof: Since the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a rhombus.

Thus, AB=BC=CD=DA

In $\mathrm{△}$ BAD and $\mathrm{△}$ ABC,

AD=BC (proved above )

AB=AB (common)

BD=AC

$\mathrm{△}$ BAD $\cong$ $\mathrm{△}$ ABC (By SSS)

$\mathrm{\angle }BAD=\mathrm{\angle }ABC$ (CPCT)

$\mathrm{\angle }$ BAD+ $\mathrm{\angle }$ ABC = ${180}^{\circ }$ (Co-interior angles)

2. $\mathrm{\angle }$ ABC = ${180}^{\circ }$

$\mathrm{\angle }$ ABC = ${90}^{\circ }$

Hence, the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.

Q6 (i) Diagonal AC of a parallelogram ABCD bisects $\mathrm{\angle }A$ (see Fig. $8.19$ ). Show that

it bisects $\mathrm{\angle }C$ also.

Answer:

Given: $\mathrm{\angle }$ DAC= $\mathrm{\angle }$ BAC ................1

$\mathrm{\angle }$ DAC= $\mathrm{\angle }$ BCA.................2 (Alternate angles)

$\mathrm{\angle }$ BAC= $\mathrm{\angle }$ ACD .................3 (Alternate angles)

From equation 1,2 and 3, we get

$\mathrm{\angle }$ ACD= $\mathrm{\angle }$ BCA...................4

Hence, diagonal AC bisect angle C also.

Q6 (ii) Diagonal AC of a parallelogram ABCD bisects $\mathrm{\angle }A$ (see Fig. $8.19$ ). Show that

ABCD is a rhombus.

Answer:

Given: $\mathrm{\angle }$ DAC= $\mathrm{\angle }$ BAC ................1

$\mathrm{\angle }$ DAC= $\mathrm{\angle }$ BCA.................2 (Alternate angles)

$\mathrm{\angle }$ BAC= $\mathrm{\angle }$ ACD .................3 (Alternate angles)

From equation 1,2 and 3, we get

$\mathrm{\angle }$ ACD= $\mathrm{\angle }$ BCA...................4

From 2 and 4, we get

$\mathrm{\angle }$ ACD= $\mathrm{\angle }$ DAC

In $\mathrm{△}$ ADC,

$\mathrm{\angle }$ ACD= $\mathrm{\angle }$ DAC (proved above )

AD=DC (In a triangle,sides opposite to equal angle are equal)

A parallelogram whose adjacent sides are equal , is a rhombus.

Thus, ABCD is a rhombus.

Q7 ABCD is a rhombus. Show that diagonal AC bisects $\mathrm{\angle }A$ as well as $\mathrm{\angle }C$ and diagonal BD bisects $\mathrm{\angle }B$ as well as $\mathrm{\angle }D$ .

Answer:

In $\mathrm{△}$ ADC,

AD = CD (ABCD is a rhombus)

$\mathrm{\angle }$ 3= $\mathrm{\angle }$ 1.................1(angles opposite to equal sides are equal )

$\mathrm{\angle }$ 3= $\mathrm{\angle }$ 2.................2 (alternate angles)

From 1 and 2, we have

$\mathrm{\angle }$ 1= $\mathrm{\angle }$ 2.................3

and $\mathrm{\angle }$ 1= $\mathrm{\angle }$ 4.................4 (alternate angles)

From 1 and 4, we get

$\mathrm{\angle }$ 3= $\mathrm{\angle }$ 4.................5

Hence, from 3 and 5, diagonal AC bisects angles A as well as angle C.

In $\mathrm{△}$ ADB,

AD = AB (ABCD is a rhombus)

$\mathrm{\angle }$ 5= $\mathrm{\angle }$ 7.................6(angles opposite to equal sides are equal )

$\mathrm{\angle }$ 7= $\mathrm{\angle }$ 6.................7 (alternate angles)

From 6 and 7, we have

$\mathrm{\angle }$ 5= $\mathrm{\angle }$ 6.................8

and $\mathrm{\angle }$ 5= $\mathrm{\angle }$ 8.................9(alternate angles)

From 6 and 9, we get

$\mathrm{\angle }$ 7= $\mathrm{\angle }$ 8.................10

Hence, from 8 and 10, diagonal BD bisects angles B as well as angle D.

Q8 (i) ABCD is a rectangle in which diagonal AC bisects $\mathrm{\angle }A$ as well as $\mathrm{\angle }C$ . Show that:

ABCD is a square

Answer:

Given: ABCD is a rectangle with AB=CD and BC=AD $\mathrm{\angle }$ 1= $\mathrm{\angle }$ 2 and $\mathrm{\angle }$ 3= $\mathrm{\angle }$ 4.

To prove: ABCD is a square.

Proof : $\mathrm{\angle }$ 1= $\mathrm{\angle }$ 4 .............1(alternate angles)

$\mathrm{\angle }$ 3= $\mathrm{\angle }$ 4 ................2(given )

From 1 and 2, $\mathrm{\angle }$ 1= $\mathrm{\angle }$ 3.....................................3

In $\mathrm{△}$ ADC,

$\mathrm{\angle }$ 1= $\mathrm{\angle }$ 3 (from 3 )

DC=AD (In a triangle, sides opposite to equal angle are equal )

A rectangle whose adjacent sides are equal is a square.

Hence, ABCD is a square.

Q8 (ii) ABCD is a rectangle in which diagonal AC bisects $\mathrm{\angle }A$ as well as $\mathrm{\angle }C$ . Show that:

diagonal BD bisects $\mathrm{\angle }B$ as well as $\mathrm{\angle }D$ .

Answer:

In $\mathrm{△}$ ADB,

AD = AB (ABCD is a square)

$\mathrm{\angle }$ 5= $\mathrm{\angle }$ 7.................1(angles opposite to equal sides are equal )

$\mathrm{\angle }$ 5= $\mathrm{\angle }$ 8.................2 (alternate angles)

From 1 and 2, we have

$\mathrm{\angle }$ 7= $\mathrm{\angle }$ 8.................3

and $\mathrm{\angle }$ 7 = $\mathrm{\angle }$ 6.................4(alternate angles)

From 1 and 4, we get

$\mathrm{\angle }$ 5= $\mathrm{\angle }$ 6.................5

Hence, from 3 and 5, diagonal BD bisects angles B as well as angle D.

Q9 (i) In parallelogram ABCD, two points P and Q are taken on diagonal BD such that $DP=BQ$ (see Fig. $8.20$ ). Show that: $\mathrm{\Delta }APD\cong \mathrm{\Delta }CQB$

Answer:

Given: In parallelogram ABCD, two points P and Q are taken on diagonal BD such that $DP=BQ$ .

To prove : $\mathrm{\Delta }APD\cong \mathrm{\Delta }CQB$

Proof :

In $\mathrm{\Delta }APDand\mathrm{\Delta }CQB,$

DP=BQ (Given )

$\mathrm{\angle }$ ADP= $\mathrm{\angle }$ CBQ (alternate angles)

AD=BC (opposite sides of a parallelogram)

$\mathrm{\Delta }APD\cong \mathrm{\Delta }CQB$ (By SAS)

Q9 (ii) In parallelogram ABCD, two points P and Q are taken on diagonal BD such that $DP=BQ$ (see Fig. $8.20$ ). Show that: $AP=CQ$

Answer:

Given: In parallelogram ABCD, two points P and Q are taken on diagonal BD such that $DP=BQ$ .

To prove : $AP=CQ$

Proof :

In $\mathrm{\Delta }APDand\mathrm{\Delta }CQB,$

DP=BQ (Given )

$\mathrm{\angle }$ ADP= $\mathrm{\angle }$ CBQ (alternate angles)

AD=BC (opposite sides of a parallelogram)

$\mathrm{\Delta }APD\cong \mathrm{\Delta }CQB$ (By SAS)

$AP=CQ$ (CPCT)

Q9 (iii) In parallelogram ABCD, two points P and Q are taken on diagonal BD such that $DP=BQ$ (see Fig. $8.20$ ). Show that: $\mathrm{\Delta }AQB\cong \mathrm{\Delta }CPD$

Answer:

Given: In parallelogram ABCD, two points P and Q are taken on diagonal BD such that $DP=BQ$ .

To prove : $\mathrm{\Delta }AQB\cong \mathrm{\Delta }CPD$

Proof :

In $\mathrm{\Delta }AQBand\mathrm{\Delta }CPD,$

DP=BQ (Given )

$\mathrm{\angle }$ ABQ= $\mathrm{\angle }$ CDP (alternate angles)

AB=CD (opposite sides of a parallelogram)

$\mathrm{\Delta }AQB\cong \mathrm{\Delta }CPD$ (By SAS)

Q9 (iv) In parallelogram ABCD, two points P and Q are taken on diagonal BD such that $DP=BQ$ (see Fig. $8.20$ ). Show that: $AQ=CP$

Answer:

Given: In parallelogram ABCD, two points P and Q are taken on diagonal BD such that $DP=BQ$ .

To prove : $AQ=CP$

Proof :

In $\mathrm{\Delta }AQBand\mathrm{\Delta }CPD,$

DP=BQ (Given )

$\mathrm{\angle }$ ABQ= $\mathrm{\angle }$ CDP (alternate angles)

AB=CD (opposite sides of a parallelogram)

$\mathrm{\Delta }AQB\cong \mathrm{\Delta }CPD$ (By SAS)

$AQ=CP$ (CPCT)

Q9 (v) In parallelogram ABCD, two points P and Q are taken on diagonal BD such that $DP=BQ$ (see Fig. $8.20$ ). Show that: APCQ is a parallelogram

Answer:

Given: In parallelogram ABCD, two points P and Q are taken on diagonal BD such that $DP=BQ$ .

To prove: APCQ is a parallelogram

Proof :

In $\mathrm{\Delta }APDand\mathrm{\Delta }CQB,$

DP=BQ (Given )

$\mathrm{\angle }$ ADP= $\mathrm{\angle }$ CBQ (alternate angles)

AD=BC (opposite sides of a parallelogram)

$\mathrm{\Delta }APD\cong \mathrm{\Delta }CQB$ (By SAS)

$AP=CQ$ (CPCT)...............................................................1

Also,

In $\mathrm{\Delta }AQBand\mathrm{\Delta }CPD,$

DP=BQ (Given )

$\mathrm{\angle }$ ABQ= $\mathrm{\angle }$ CDP (alternate angles)

AB=CD (opposite sides of a parallelogram)

$\mathrm{\Delta }AQB\cong \mathrm{\Delta }CPD$ (By SAS)

$AQ=CP$ (CPCT)........................................2

From equation 1 and 2, we get

$AP=CQ$

$AQ=CP$

Thus, opposite sides of quadrilateral APCQ are equal so APCQ is a parallelogram.

Q10 (i) ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD (see Fig. $8.21$ ). Show that $\mathrm{\Delta }APB\cong \mathrm{\Delta }CQD$

Answer:

Given: ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD.

To prove : $\mathrm{\Delta }APB\cong \mathrm{\Delta }CQD$

Proof: In $\mathrm{\Delta }APBand\mathrm{\Delta }CQD$ ,

$\mathrm{\angle }$ APB= $\mathrm{\angle }$ CQD (Each ${90}^{\circ }$ )

$\mathrm{\angle }$ ABP= $\mathrm{\angle }$ CDQ (Alternate angles)

AB=CD (Opposite sides of a parallelogram )

Thus, $\mathrm{\Delta }APB\cong \mathrm{\Delta }CQD$ (By SAS)

Q10 (ii) ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD (see Fig. $8.21$ ). Show that $AP=CQ$

Answer:

Given: ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD.

To prove : $AP=CQ$

Proof: In $\mathrm{\Delta }APBand\mathrm{\Delta }CQD$ ,

$\mathrm{\angle }$ APB= $\mathrm{\angle }$ CQD (Each ${90}^{\circ }$ )

$\mathrm{\angle }$ ABP= $\mathrm{\angle }$ CDQ (Alternate angles)

AB=CD (Opposite sides of a parallelogram )

Thus, $\mathrm{\Delta }APB\cong \mathrm{\Delta }CQD$ (By SAS)

$AP=CQ$ (CPCT)

Q11 (i) In $\mathrm{\Delta }ABC$ and $\mathrm{\Delta }DEF$ , $AB=DE,AB\parallel DE,BC=EF$ and $BC\parallel EF$ . Vertices A, B and C are joined to vertices D, E and F respectively (see Fig. $8.22$ ). Show that quadrilateral ABED is a parallelogram

Answer:

Given : In $\mathrm{\Delta }ABC$ and $\mathrm{\Delta }DEF$ , $AB=DE,AB\parallel DE,BC=EF$ and $BC\parallel EF$ .

To prove : quadrilateral ABED is a parallelogram

Proof : In ABED,

AB=DE (Given)

AB||DE (Given )

Hence, quadrilateral ABED is a parallelogram.

Q11 (ii) In $\mathrm{\Delta }ABC$ and $\mathrm{\Delta }DEF$ , $AB=DE,AB\parallel DE,BC=EF$ and $BC\parallel EF$ . Vertices A, B and C are joined to vertices D, E and F respectively (see Fig. $8.22$ ). Show that quadrilateral BEFC is a parallelogram.

Answer:

Given: In $\mathrm{\Delta }ABC$ and $\mathrm{\Delta }DEF$ , $AB=DE,AB\parallel DE,BC=EF$ and $BC\parallel EF$ .

To prove: quadrilateral BEFC is a parallelogram

Proof: In BEFC,

BC=EF (Given)

BC||EF (Given )

Hence, quadrilateral BEFC is a parallelogram.

Q11 (iii) In $\mathrm{\Delta }ABC$ and $\mathrm{\Delta }DEF$ , $AB=DE,AB\parallel DE,BC=EF$ and $BC\parallel EF$ . Vertices A, B and C are joined to vertices D, E and F respectively (see Fig. $8.22$ ). Show that $AD\parallel CF$ and $AD=CF$

Answer:

To prove: $AD\parallel CF$ and $AD=CF$

Proof :

In ABED,

AD=BE.................1(ABED is a parallelogram)

AD||BE .................2(ABED is a parallelogram)

In BEFC,

BE=CF.................3(BEFC is a parallelogram)

BE||CF .................4(BEFC is a parallelogram)

From 2 and 4 , we get

AD||CF

From 1 and 3, we get

AD=CF

Q11 (iv) In $\mathrm{\Delta }ABC$ and $\mathrm{\Delta }DEF$ , $AB=DE,AB\parallel DE,BC=EF$ and $BC\parallel EF.$ . Vertices A, B and C are joined to vertices D, E and F respectively (see Fig. $8.22$ ). Show that quadrilateral ACFD is a parallelogram.

Answer:

To prove : quadrilateral ACFD is a parallelogram

Proof :

In ABED,

AD=BE.................1(ABED is a parallelogram)

AD||BE .................2(ABED is a parallelogram)

In BEFC,

BE=CF.................3(BEFC is a parallelogram)

BE||CF .................4(BEFC is a parallelogram)

From 2 and 4 , we get

AD||CF...........................5

From 1 and 3, we get

AD=CF...........................6

From 5 and 6, we get

AD||CF and AD=CF

Thus, quadrilateral ACFD is a parallelogram

Q11 (v) In $\mathrm{\Delta }ABC$ and $\mathrm{\Delta }DEF$ , $AB=DE,AB\parallel DE,BC=EF$ and $BC\parallel EF$ . Vertices A, B and C are joined to vertices D, E and F respectively (see Fig. $8.22$ ). Show that $AC=DF$

Answer:

In ACFD,

AC=DF (Since, ACFD is a parallelogram which is prooved in part (iv) of the question)

Q11 (vi) In $\mathrm{\Delta }ABC$ and $\mathrm{\Delta }DEF$ , $AB=DE,AB\parallel DE,BC=EF$ and $BC\parallel EF.$ Vertices A, B and C are joined to vertices D, E and F respectively (see Fig. $8.22$ ). Show that $\mathrm{\Delta }ABC\cong \mathrm{\Delta }DEF$

Answer:

In $\mathrm{\Delta }ABC$ and $\mathrm{\Delta }DEF$ ,

AB=DE (Given )

BC=EF (Given )

AC=DF ( proved in (v) part)

$\mathrm{\Delta }ABC\cong \mathrm{\Delta }DEF$ (By SSS rule)

Q12 (i) ABCD is a trapezium in which $AB\parallel CD$ and $AD=BC$ (see Fig. $8.23$ ). Show that $\mathrm{\angle }A=\mathrm{\angle }B$

[ Hint: Extend AB and draw a line through C parallel to DA intersecting AB produced at E.]

Answer:

Given: ABCD is a trapezium in which $AB\parallel CD$ and $AD=BC$

To prove : $\mathrm{\angle }A=\mathrm{\angle }B$

Proof: Let $\mathrm{\angle }$ A be $\mathrm{\angle }$ 1, $\mathrm{\angle }$ ABC be $\mathrm{\angle }$ 2, $\mathrm{\angle }$ EBC be $\mathrm{\angle }$ 3, $\mathrm{\angle }$ BEC be $\mathrm{\angle }$ 4.

In AECD,

AE||DC (Given)

AD||CE (By cnstruction)

Hence, AECD is a parallelogram.

AD=CE...............1(opposite sides of a parallelogram)

AD=BC.................2(Given)

From 1 and 2, we get

CE=BC

In $\mathrm{△}$ BCE,

$\mathrm{\angle }3=\mathrm{\angle }4$ .................3 (opposite angles of equal sides)

$\mathrm{\angle }2+\mathrm{\angle }3={180}^{\circ }$ ...................4(linear pairs)

$\mathrm{\angle }1+\mathrm{\angle }4={180}^{\circ }$ .....................5(Co-interior angles)

From 4 and 5, we get

$\mathrm{\angle }2+\mathrm{\angle }3=\mathrm{\angle }1+\mathrm{\angle }4$

$\therefore \mathrm{\angle }2=\mathrm{\angle }1\to \mathrm{\angle }B=\mathrm{\angle }A$ (Since, $\mathrm{\angle }3=\mathrm{\angle }4$ )

Q12 (ii) ABCD is a trapezium in which AB || CD and AD = BC (see Fig. 8.23). Show that $\mathrm{\angle }C=\mathrm{\angle }D$

[ Hint : Extend AB and draw a line through C parallel to DA intersecting AB produced at E.]

Answer:

Given: ABCD is a trapezium in which $AB\parallel CD$ and $AD=BC$

To prove : $\mathrm{\angle }C=\mathrm{\angle }D$

Proof: Let $\mathrm{\angle }$ A be $\mathrm{\angle }$ 1, $\mathrm{\angle }$ ABC be $\mathrm{\angle }$ 2, $\mathrm{\angle }$ EBC be $\mathrm{\angle }$ 3, $\mathrm{\angle }$ BEC be $\mathrm{\angle }$ 4.

$\mathrm{\angle }1+\mathrm{\angle }D={180}^{\circ }$ (Co-interior angles)

$\mathrm{\angle }2+\mathrm{\angle }C={180}^{\circ }$ (Co-interior angles)

$\therefore \mathrm{\angle }1+\mathrm{\angle }D=\mathrm{\angle }2+\mathrm{\angle }C$

Thus, $\mathrm{\angle }C=\mathrm{\angle }D$ (Since , $\mathrm{\angle }1=\mathrm{\angle }2$ )

Q12 (iii) ABCD is a trapezium in which $AB\parallel CD$ and $AD=BC$ (see Fig. $8.23$ ). Show that $\mathrm{\Delta }ABC\cong \mathrm{\Delta }BAD$

[ Hint : Extend AB and draw a line through C parallel to DA intersecting AB produced at E.]

Answer:

Given: ABCD is a trapezium in which $AB\parallel CD$ and $AD=BC$

To prove : $\mathrm{\Delta }ABC\cong \mathrm{\Delta }BAD$

Proof: In $\mathrm{\Delta }ABCand\mathrm{\Delta }BAD$ ,

BC=AD (Given )

AB=AB (Common )

$\mathrm{\angle }ABC=\mathrm{\angle }BAD$ (proved in (i) )

Thus, $\mathrm{\Delta }ABC\cong \mathrm{\Delta }BAD$ (By SAS rule)

Q12 (iv) ABCD is a trapezium in which $AB\parallel CD$ and $AD=BC$ (see Fig. $8.23$ ). Show that diagonal AC $=$ diagonal BD

[ Hint : Extend AB and draw a line through C parallel to DA intersecting AB produced at E.]

Answer:

Given: ABCD is a trapezium in which $AB\parallel CD$ and $AD=BC$

To prove: diagonal AC $=$ diagonal BD

Proof: In $\mathrm{\Delta }ABCand\mathrm{\Delta }BAD$ ,

BC=AD (Given )

AB=AB (Common )

$\mathrm{\angle }ABC=\mathrm{\angle }BAD$ (proved in (i) )

Thus, $\mathrm{\Delta }ABC\cong \mathrm{\Delta }BAD$ (By SAS rule)

diagonal AC $=$ diagonal BD (CPCT)