Question

$ABCD$ is a trapezium in which $AB||CD$ and $AD=BC$. Show that $\angle A=\angle B$.

Answer 1

Given that ABCD is a trapezium,

$AB\parallel CD$ and $AD=BC$

To prove: $\angle A=\angle B$

Construction: Draw $CE\parallel AD$ and extend AB to intersect CE at E.

Poof: As AECD is a parallelogram.

By constriction, CE is parallel to AD and AE is parallel to CD,

$\therefore AD=EC$

But, AD=BC (Given)

$\therefore BC=EC$

$\Rightarrow \angle 3=\angle 4$ (Angle opposite to equal side are equal)

Now, $\angle 1+\angle 4={180}^0$ (consecutive interior angle of parallelogram )

And $\angle 2+\angle 3={180}^0$ (linear pair )

$\Rightarrow \angle 1+\angle 4=\angle 2+\angle 3$ $\Rightarrow \angle 1=\angle 2\because \angle 3=\angle 4$ (so get cancelled with each other )

$\Rightarrow \angle A=\angle B$

Hence, proved .