Question

In the given figure, AB || CD. Prove that ∠BAE − ∠DCE = ∠AEC.

Answer 1

Draw $EF\parallel AB\parallel CD$ through E.
Now, $EF\parallel AB$ and AE is the transversal.
Then, $\angle BAE+\angle AEF=180°\left[\text{Angles on the same side of a transversal line are supplementary}\right]$
Again, $EF\parallel CD$ and CE is the transversal.
Then,
$$\begin{gathered} \angle DCE+\angle CEF=180°\left[\text{Angles on the same side of a transversal line are supplementary}\right] \\ \Rightarrow \angle DCE+\left(\angle AEC+\angle AEF\right)=180° \\ \Rightarrow \angle DCE+\angle AEC+180°-\angle BAE=180° \\ \Rightarrow \angle BAE-\angle DCE=\angle AEC \end{gathered}$$