ABCD is a trapezium in which AB || CD and AD = BC. Show that :

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MCQ

$\text{ABCD}$ is a trapezium in which $AB || CD$ and $AD = BC$.
Show that :

A
$\angle A = \angle B$
B
$\angle C = \angle D$
C
$\triangle ABC \cong \angle BAD$
✓
diagonal $AC =$ diagonal $BD.$

✓

Answer

Correct option: D.

diagonal $AC =$ diagonal $BD.$

Given : $\text{ABCD}$ is a trapezium in which $AB || CD$ and $AD = BC.$
To Prove :

$\angle A = \angle B$
$\angle C = \angle D$
$\triangle ABC \cong \triangle BAD$
diagonal $AC =$ diagonal $BD$.

Construction: Extend $AB$ and draw a line through $C$ parallel to $DA$ intersecting $AB$ produced at $E.$

Proof :

$AB || CD . . . [$Given$]$
$AD || ED . . .[$By construction$]$
$\therefore AECD$ is a parallelogram $. . . [$A quadrilateral is a parallelogram if a pair of Opp. sides are parallel and of equal length$]$
$\therefore AD = EC . . .[$Opp. sides of a $||\ gm$ are equal$]$
But $AD = BC . . .[$Given$]$
$\therefore EC = BC$
$\therefore \angle CBE = \angle CEB . . . [ \angle$s opposite to equal side of a triangle are equal$] . . .(1)$
$\angle B + \angle CBE = 180^\circ . . .[$Linear pair axiom$]. . . (2)$
As $AD || EC . . .[$By construction$]$
and transversal $AE$ intersect them
$\angle A + \angle CEB = 180^\circ . . . [$The sum of interior angles on the same side of the transversal is $180^\circ] . . . (3)$
From $(2)$ and $(3),$
$\angle B + \angle CBE = \angle A + \angle CEB$
But $\angle CBE = \angle CEB . . .[$ From $(1)]$
$\therefore \angle B = \angle A$ or $\angle A = \angle B$
As $AB || CD$
$\therefore \angle A + \angle D = 180^\circ. . .[$The sum of interior angles on the same side of the transversal is $180^\circ]$
and $\angle B + \angle C = 180^\circ$
$\therefore \angle A + \angle D = \angle B + \angle C$
But $\angle A = \angle B . . .[$As proved in $(i)]$
$\therefore \angle D = \angle C$ or $\angle C = \angle D$
In $\triangle ABC$ and $\triangle BAD,$
$AB = BA . . . [$Common$]$
$BC = AD . . .[$Given$]$
$\angle ABC = \angle BAD . . .[$ From $(i)]$
$\therefore \triangle ABC \cong \triangle BAD . . .[\text{SAS}$ rule$]$
As $\triangle ABC \cong \triangle BAD . . [$From $(iii)]$
$\therefore AC = BD . . .[c.p.c.t.]$