Question

$P$ is a point on the bisector of $\angle ABC$. The line through $P$ parallel to $BA$ meets $BC$ at $Q$. Prove that $\Delta BPQ$ is an isosceles triangle.

Answer 1

$\angle ABP=\angle PBC$

From the figure,

$\therefore \angle 3=\angle 2$

$⟹PQ||BC$

$\therefore \angle 1=\angle 2$ ....... (Alternate interior angles)

$\angle 1=\angle 3\{\angle 2=\angle 3\}$

In on isosceles triangle and the base are same.

$\angle 1=\angle 3$

$\therefore △BPQ$ is an isosceles triangle.