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

Question:

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

 

Solution:

Draw $P F Q\|A B\| C D$.

Now, $P F Q \| A B$ and EF is the transversal.

Then,

$\angle A E F+\angle E F P=180^{\circ} \ldots \ldots(1)$

[Angles on the same side of a transversal line are supplementary]

Also, $P F Q \| C D$.

$\angle P F G=\angle F G D=r^{\circ}[$ 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 eqn. (i)

we get,

$p^{\circ}+q^{\circ}-r^{\circ}=180^{\circ}$

$\Rightarrow p+q-r=180$

 

 

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