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]

$\Rightarrow x^{\circ}+20^{\circ}+50^{\circ}=180^{\circ} \quad[\angle A E F=\angle A E C+\angle C E F]$

$\Rightarrow x^{\circ}+70^{\circ}=180^{\circ}$

$\Rightarrow x^{\circ}=110^{\circ}$

 

$\Rightarrow x=110$