Question

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

Answer 1

Given.

$BP$ is the bisector of $\angle ABC$.

the line through $P$, parallel to $BA$meet $BC$ at $Q$

To Prove.

$BPQ$ is an isosceles triangle.

Proof.

From the given information, a diagram is drawn which is shown below.

Since, $BP$ is the bisector of $\angle ABC$,

$\angle 1=\angle 2\dots \left(1\right)$

and $PQ$ is parallel to $BA$.

$\therefore \angle 1=\angle 3\dots \left(2\right)[\mathrm{Alternate}\mathrm{angles}]$

From equations, $\left(1\right)\mathrm{and}\left(2\right)$,

$\therefore \angle 2=\angle 3$

In $\Delta BPQ$,

$$\begin{gathered} \because \angle 2=\angle 3 \\ \therefore PQ=BQ\left[\mathrm{if}\mathrm{two}\mathrm{angles}\mathrm{of}\mathrm{a}\mathrm{triangle}\mathrm{are}\mathrm{equal},\mathrm{then}\mathrm{opposite}\mathrm{sides}\mathrm{to}\mathrm{them}\mathrm{are}\mathrm{also}\mathrm{equal}.\right] \end{gathered}$$

Hence, $BPQ$ is an isosceles triangle.