Question

In figure AB || DC and AD || BC. Prove that $\angle DAB=\angle DCB.$

Answer 1

Since AD || BC and AB is a transversal intersecting them at A and B respectively.
Therefore
$\angle DAB+\angle ABC={108}^{\circ }$ ....(i) [Consecutive interior angles]
Again, AB || CD and BC is a transversal intersecting them at B and C respectively.
Therefore,
$\angle ABC+\angle DCB={108}^{\circ }$ ...(ii) [Consecutive interior angles]
From (i) and (ii), we get
$\angle DAB+\angle ABC=\angle ABC+\angle DCB$
$\Rightarrow \angle DAB=\angle DCB$