Question

The figure shows two parallel lines AB and DE cut by the transversals AE and BD:

Answer 1

Answer: $\triangle ABC\sim \triangle EDC$

That is, triangles ABC and EDC are similar.

Step-by-step explanation:

Given : $AB\parallel DE$

AE and BD are the common transversal of the parallel lines AB and DE,

Then By the alternative interior angle theorem,

$\angle BAC\cong \angle DEC$

And, $\angle ABC\cong \angle EDC$

Thus, By AA similarity postulate,

$\triangle ABC\sim \triangle EDC$

⇒ triangles ABC and EDC are similar.  

Answer 2

I'm guessing it's the angles, for example: 3 is congruent to 4; and D is congruent to A, etc.