Question

If two parallel lines are intersected by a transversal, prove that the bisectors of the interior angles on the same side of transversal intersect each other at right angles.

Answer 1

We know that the sum of the interior angles on the same side of transversal is ${180}^{\circ }$.

Therefore, $\angle$BMN + $\angle$DNM = ${180}^{\circ }$

$\Rightarrow$ $\frac{1}{2}$$\angle$BMN + $\frac{1}{2}$ $\angle$DNM = ${90}^{\circ }$

$\angle$1 + $\angle$2 = ${90}^{\circ }$ ...(i)

Now, In $△$PMN, we have

$\angle$1 + $\angle$2 + $\angle$3 = ${180}^{\circ }$ ....(ii)

From (i) and (ii), we get

$\angle$3 = ${90}^{\circ }$

$\Rightarrow$ PM and PN intersect at right angles.