Question

P is a point on the bisector of an angle $\angle ABC.$ If the line through P parallel to AB meets BC at Q, prove that triangle BPQ is isosceles.

Answer 1

Given : In $\Delta ABC,P\text{is a point on the bisector of}\angle Bandfrom,P,RPQ||AB\text{is drawn which meets BC in Q}$

To prove : $\Delta BPQ$ is an isosceles

Proof : $\because$ BD is the bisectors of CB

$\therefore \angle 1=\angle 2$

$\because RPQ||AB$

$\therefore \angle 1=\angle 3\left(\right(\text{Alternate angles)}$

But $\angle 1=\angle 2\left(Proved\right)$

$\therefore \angle 2=\angle 3$

$\therefore$ $PQ=BQ$ (Sides opposite to equal angles)

$\therefore \Delta BPQ$ is an isosceles