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

Draw $E F\|A B\| C D$ through $\mathrm{E}$.

Now, $E F \| A B$ and $\mathrm{AE}$ is the transversal.

Then, $\angle B A E+\angle A E F=180^{\circ} \quad$ [Angles on the same side of a transversal line are supplementary]

Again, $E F \| C D$ and $\mathrm{CE}$ is the transversal.

Then

$\angle D C E+\angle C E F=180^{\circ} \quad$ [Angles on the same side of a transversal line are supplementary]

$\Rightarrow \angle D C E+(\angle A E C+\angle A E F)=180^{\circ}$

$\Rightarrow \angle D C E+\angle A E C+180^{\circ}-\angle B A E=180^{\circ}$

$\Rightarrow \angle B A E-\angle D C E=\angle A E C$