In the given figure, if $A B \| D E, \angle B A C=35^{\circ}$ and $\angle C D E=53^{\circ}$, find $\angle D C E$.

Solution:

$A B \| D E$ and $A E$ is a transversal.

$\angle B A C=\angle C E D$ (Alternate interior angles)

$\therefore \angle C E D=35^{\circ}$

In $\triangle \mathrm{CDE}$

$\angle C D E+\angle C E D+\angle D C E=180^{\circ}$ (Angle sum property of a triangle)

$53^{\circ}+35^{\circ}+\angle D C E=180^{\circ}$

$\angle \mathrm{DCE}=180^{\circ}-88^{\circ}$

$\angle \mathrm{DCE}=92^{\circ}$