Question
In the given figure, $BAD\ ||\ EF$, $\angle\text{AEF}=55^\circ$ and $\angle\text{ACB}=25^\circ,$ find $\angle\text{ABC}.$

Answer

$BAD\ ||\ EF$ and $EC$ is the transversal.
$\Rightarrow\angle\text{AEF}=\angle\text{CAD}$ (corresponding angles)
$\Rightarrow\angle\text{CAD}=55^\circ$ Now,
$\angle\text{CAD}+\angle\text{CAB}=180^\circ$ (linear pair)
$\Rightarrow55^\circ+\angle\text{CAB}=180^\circ$
$\Rightarrow\angle\text{CAB}=125^\circ$ In $\triangle\text{ABC},$
by angle sum property, $\angle\text{ABC}+\angle\text{CAB}+\angle\text{ACB}=180^\circ$
$\Rightarrow\angle\text{ABC}+125^\circ+25^\circ=180^\circ$
$ \Rightarrow\angle\text{ABC}=30^\circ$

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