Question

Question 12 (iii)
ABCD is a trapezium in which AB | | CD and AD =BC (see the given figure). Show that \(\Delta ABC \cong \Delta BAD\).

Answer 1

Extend $AB$ such that it intersects $CE$ at point $E$ where $CE\parallel AD$.

Since, $AB\parallel CD$ and $CE\parallel AD$, therefore, $AECD$ is a parallelogram.

Since, $ABCD$ is a trapezium, therefore,

$\Rightarrow \angle ABC+\angle ADC={180}^{\circ }$

$\Rightarrow \angle ABC={180}^{\circ }-\theta$ ...(i)

Also,

$\Rightarrow \angle ABC+\angle CBE={180}^{\circ }$ [Linear Pair]

$\therefore \angle CBE=\theta$ ...(ii)

Since, $AECD$ is a parallelogram, therefore, $\angle BEC=\angle ADC=\theta$ ...(iii)

From (ii) and (iii)

$\Rightarrow \angle BEC=\angle CBE=\theta$

Since, $AE\parallel DC$

Therefore, $\angle DAE={180}^{\circ }-\theta$...(iv) [Angles on same side of transversal are supplementary]

Now, in $\Delta ABC$ and $\Delta BAD$.

$AD=BC$ [Given]

$AB=AB$ (Common)

$\angle DAB=\angle ABC={180}^{\circ }-\theta$ [From (i) and (iv)]

$\Delta ABC\cong \Delta BAD$ by SAS rule.