Prove that AB ≅ DC

The correct answer is: AB = DC (corresponding parts of congruent triangles)

Complete step by step solution:
Consider 2 triangles, △ AEB and △ CED
From the figure, we have
AE = CE (given)
∠ ABE = ∠ CDE = 55° (alternate angles)
⇒ AB ∥ CD
So, we have ∠EAB=∠ECD=65° (alternate angles, since )
Now, ∠ CED = 180°- (65° + 55°) =180° - 120° = 60° (angle sum property)
∠ CED = ∠ AEB = 60° (vertically opposite angles)
In △ AEB and △ CED,
∠ EAB=∠ECD=65°(alternate angles)
AE = CE (given)
∠ CED = ∠ AEB = 60° (vertically opposite angles)
So, △ AEB ⩭ △ CED by ASA congruence criterion
⇒ AB = DC (corresponding parts of congruent triangles)