Question

In the given figure ,AB parallel to CD . Prove that angle BAE- angle ECD = AEC​

Answer 1

Answer:

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

Then,

∠AFE=∠ECD

Also, ∠BAE=∠AFE+∠AEC

Therefore,

∠BAE=∠ECD+∠AEC

∠BAE−∠ECD=∠AEC