In the given figure, AB || CD. If ∠EAB = 50° and ∠ECD = 60°, then ∠AEB = ?

(c) 70°

$A B \| C D$ and $B C$ is the transversal.

$\therefore \angle A B E=\angle B C D=60^{\circ} \quad[$ Alternate Internal Angles $]$

In $\Delta A B E$, we have:

$\angle E A B+\angle A B E+\angle A E B=180^{\circ} \quad[$ Sum of the angles of a triangle $]$

$\Rightarrow 50^{\circ}+60^{\circ}+\angle A E B=180^{\circ}$

$\Rightarrow \angle A E B=70^{\circ}$