Question

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

Answer 1

Draw $EF\parallel AB\parallel CD$ through E.
Now, $EF\parallel AB$ and AE is the transversal.
Then, ∠BAE+∠AEF=180°

Angles on the same side of a transversal line are supplementary
Again, $EF\parallel CD$ and CE is the transversal.
Then,
∠DCE+∠CEF=180° (Angles on the same side of a transversal line are supplementary)

⇒∠DCE+∠AEC+∠AEF=180°
⇒∠DCE+∠AEC+180°-∠BAE=180°

⇒∠BAE-∠DCE=∠AEC