Question

In the given figure, $AB||CD.$ Prove that $\angle BAE-\angle ECD=\angle AEC$

Answer 1

Produce BA to touch EC. Let the point of intersection be F.

Then,

$\angle AFE=\angle ECD$

Also, $\angle BAE=\angle AFE+\angle AEC$

Therefore,

$\angle BAE=\angle ECD+\angle AEC$

$\angle BAE-\angle ECD=\angle AEC$