Question

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

Answer 1

Draw $PFQ\parallel AB\parallel CD$.
Now, $PFQ\parallel AB$ and EF is the transversal.
Then,
$$\begin{gathered} \angle AEF+\angle EFP=180°.....\left(1\right) \\ \left[\text{Angles on the same side of a transversal line are supplementary}\right] \end{gathered}$$
Also, $PFQ\parallel CD$.

$$\begin{gathered} \angle PFG=\angle FGD=r°\left[\text{Alternate Angles}\right] \\ \mathrm{and}\angle EFP=\angle EFG-\angle PFG=q°-r° \\ \mathrm{putting}\mathrm{the}\mathrm{value}\mathrm{of}\angle EFP\mathrm{in}\mathrm{eqn}.\left(\mathrm{i}\right) \\ \mathrm{we}\mathrm{get}, \\ p°+q°-r°=180° \\ \Rightarrow p+q-r=180 \end{gathered}$$