Question

P is a point on the bisector of ∠ABC. If the line through P, parallel to BA meets BC at Q, prove that ΔBPQ is an isosceles triangle.

Answer 1

Given: BP is the bisector of ∠ABC, and BA$\parallel$QP

To prove: ΔBPQ is an isosceles triangle

Proof:

$$\begin{gathered} \because \angle 1=\angle 2\left(\mathrm{Given},BP\mathrm{is}\mathrm{the}\mathrm{bisector}\mathrm{of}\angle ABC\right) \\ \mathrm{And},\angle 1=\angle 3\left(\mathrm{Alternate}\mathrm{interior}\mathrm{angles}\right) \\ \therefore \angle 2=\angle 3 \\ \mathrm{So},PQ=BQ\left(\mathrm{In}\mathrm{a}\mathrm{triangle},\mathrm{sides}\mathrm{opposite}\mathrm{to}\mathrm{equal}\mathrm{sides}\mathrm{are}\mathrm{equal}.\right) \end{gathered}$$

But these are sides of $∆BPQ$.

Hence, $∆BPQ$ is an isosceles triangle.