$ \mathrm{ABCD} $ is a trapezium in which $ \mathrm{AB} \| \mathrm{CD} $ and $ \mathrm{AD}=\mathrm{BC} $ (see below figure). Show that
(i) $ \angle \mathrm{A}=\angle \mathrm{B} $
(ii) $ \angle \mathrm{C}=\angle \mathrm{D} $
(iii) $ \triangle \mathrm{ABC} \equiv \triangle \mathrm{BAD} $
(iv) diagonal $ \mathrm{AC}= $ diagonal $ \mathrm{BD} $
[Hint: Extend $ \mathrm{AB} $ and draw a line through $ \mathrm{C} $ parallel to $ \mathrm{DA} $ intersecting $ \mathrm{AB} $ produced at E.]
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Given:

\( \mathrm{ABCD} \) is a trapezium in which \( \mathrm{AB} \| \mathrm{CD} \) and \( \mathrm{AD}=\mathrm{BC} \).

To do :

We have to show that

(i) \( \angle \mathrm{A}=\angle \mathrm{B} \)

(ii) \( \angle \mathrm{C}=\angle \mathrm{D} \)

(iii) \( \triangle \mathrm{ABC} \equiv \triangle \mathrm{BAD} \)

(iv) diagonal \( \mathrm{AC}= \) diagonal \( \mathrm{BD} \)

Solution :  

Extend \( \mathrm{AB} \) and draw a line through \( \mathrm{C} \) parallel to \( \mathrm{DA} \) intersecting \( \mathrm{AB} \) produced at $E$.

(i) $ADCE$ is a parallelogram.

$CE = AD$      (Opposite sides of a parallelogram are equal)

$AD = BC$         (Given)

This implies,

$BC = CE$

$\Rightarrow \angle CBE = \angle CEB$

$\angle EAD+\angle CEB = 180^o$

$\angle EAD+\angle CBE = 180^o$           ($\angle CBE = \angle CEB$)

$\angle CBA +\angle CBE = 180^o$            (Linear pair)

This implies,

$\angle A = \angle B$

(ii) $\angle A+\angle D =180^o$  (Adjacent angles of a parallelogram are supplementary)

$\angle B+\angle C = 180^o$ (Angles on the same side of transversal are supplementary)

$\angle A+\angle D = \angle B+\angle C$

$\angle A+\angle D = \angle A+\angle C$          ($\angle A = \angle B$)

This implies,

$\angle D = \angle C$

(iii) In $\triangle ABC$ and $\triangle BAD$,

$AB = AB$         (Common side)

$\angle DBA = \angle CBA$

$AD = BC$          (Given)

Therefore, by SAS congruency, we get,

$\triangle ABC \cong \triangle BAD$

(iv) $\triangle ABC \cong \triangle BAD$

This implies,

$AC = BD$         (CPCT)

Therefore,

diagonal \( \mathrm{AC}= \) diagonal \( \mathrm{BD} \).