Question

ABCD is a trapezium in which AB parallel CD and AD =BC (see the given figure). Show that $\angle A=\angle B$.

Answer 1

Let us extend AB. Then, draw a line through C, which is parallel to AD, intersecting AE at point E. It is clear that AECD is a parallelogram.
AD=CE (Opposite sides of parallelogram AECD)
However, AD=BC (Given)
Therefore, BC=CE
$\angle CEB=\angle CBE$
$\angle CBE+\angle CBA={180}^{\circ }$ Linear pair)………….(1)
$\angle CBA+\angle DAB=180$ (Angles on the same side of transversal)……….(2)
From (1) and (2) $\angle A=\angle B$