$P$ is a point on the bisector of $\angle ABC$. If the line through $P$,parallel to $BA$,meets $BC$ at $Q$,prove that $\triangle BPQ$ is an isosceles triangle.
(N/A) We have to prove that $\triangle BPQ$ is an isosceles triangle.
Given that $BP$ is the bisector of $\angle ABC$,we have:
$\angle 1 = \angle 2$ .....$(1)$
Since $PQ \parallel BA$ and $BP$ is a transversal,the alternate interior angles are equal:
$\angle 1 = \angle 3$ .....$(2)$
From equations $(1)$ and $(2)$,we get:
$\angle 2 = \angle 3$
In $\triangle BPQ$,since $\angle 2 = \angle 3$,the sides opposite to these equal angles must be equal:
$PQ = BQ$
Since two sides of $\triangle BPQ$ are equal,$\triangle BPQ$ is an isosceles triangle.
Given that $BP$ is the bisector of $\angle ABC$,we have:
$\angle 1 = \angle 2$ .....$(1)$
Since $PQ \parallel BA$ and $BP$ is a transversal,the alternate interior angles are equal:
$\angle 1 = \angle 3$ .....$(2)$
From equations $(1)$ and $(2)$,we get:
$\angle 2 = \angle 3$
In $\triangle BPQ$,since $\angle 2 = \angle 3$,the sides opposite to these equal angles must be equal:
$PQ = BQ$
Since two sides of $\triangle BPQ$ are equal,$\triangle BPQ$ is an isosceles triangle.
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