We have:
$\angle\text{ABC}+\angle\text{ADC}=180^\circ$
$\Rightarrow\angle\text{ABC}+95^\circ=180^\circ$
$\Rightarrow\angle\text{ABC}=(180^\circ-95^\circ)=85^\circ$
Now, $CF || AB$ and $CB$ is the transversal.
$\therefore\angle\text{BCF}=\angle\text{ABC}=85^\circ($Alternate interior angles$)$
$\Rightarrow\angle\text{BCE}=(85^\circ+20^\circ)=105^\circ$
$\Rightarrow\angle\text{DCB}=(180^\circ-105^\circ)=75^\circ$
$\Rightarrow\angle\text{DCB}=75^\circ$
Now, $\angle\text{BAD}+\angle\text{BCD}=180^\circ$
$\Rightarrow\angle\text{BAD}+75^\circ=180^\circ$
$\Rightarrow\angle\text{BAD}=(180^\circ-75^\circ)$
$\Rightarrow\angle\text{BAD}=105^\circ$
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