Question

Given a parallelogram ABCD. Prove that $\angle DCB=\angle DAB$

Answer 1

Consider triangles $ABD$ and $BCD$ :

$AB=CD$ and $AD=BC$ (since $ABCD$ is a parallelogram, opposite sides are equal)

Also, side $BD$ is common in both the triangles.

Thus, $ABD$ and $BCD$ are Congruent triangles.

Hence, $\angle DCB=\angle DAB$ (Corresponding Angles of Congruent Triangles)