Question:
In the given figures, AB || CD. Find the value of x.
Solution:
Draw $E F\|A B\| C D$.
$E F \| C D$ and CE is the transversal.
Then,
$\angle E C D+\angle C E F=180^{\circ} \quad$ [Angles on the same side of a transversal line are supplementary]
$\Rightarrow 130^{\circ}+\angle C E F=180^{\circ}$
$\Rightarrow \angle C E F=50^{\circ}$
Again, $E F \| A B$ and $\mathrm{AE}$ is the transversal.
Then,
$\angle B A E+\angle A E F=180^{\circ}$ [Angles on the same side of a transversal line are supplementary]
$\Rightarrow x^{\circ}+20^{\circ}+50^{\circ}=180^{\circ} \quad[\angle A E F=\angle A E C+\angle C E F]$
$\Rightarrow x^{\circ}+70^{\circ}=180^{\circ}$
$\Rightarrow x^{\circ}=110^{\circ}$
$\Rightarrow x=110$