In the given figure, AB | CD. Prove that p + q - r =180. - Vidyadip

Question

In the given figure, $AB \| CD$. Prove that $p + q - r =180$.

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Answer

Draw $PFQ$ $\| AB \| CD$
Now, $PFQ$ $\| AB$ and $EF$ is the transversal.
Then,
$\angle A E F+\angle E F P=180^{\circ} \ldots \text { (i) }$
[Angles on the same side of a transversal line are supplementary]
Also, $PFQ$ $\|$ $CD$.
$\angle P F G=\angle F G D=r^{\circ}$ [Alternate Angles]
and $\angle E F P=\angle E F G-\angle P F G=q^{\circ}-r^{\circ}$
putting the value of $\angle E F P$ in equation (i)
we get,
$p^{\circ}+q^{\circ}-r^{\circ}=180^{\circ}\left[\angle AEF=p^{\circ}\right]$

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