$P$ 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.
Given:
$P$ is a point on the bisector of an angle $\angle ABC$. The line through $P$ parallel to $AB$ meets $BC$ at $Q$.
To do:
We have to prove that $BPQ$ is isosceles.
Solution:
From the figure,
$BD$ is the bisector of $CB$
This implies,
$\angle 1 = \angle 2$......(i)
$RQ \parallel AB$
This implies,
$\angle 1 = \angle 3$.......(ii) (Alternate angles)
From (i) and (ii), we get,
$\angle 2 = \angle 3$
This implies,
$PQ = BQ$ (Sides opposite to equal angles are equal)
Therefore, $\triangle BPQ$ is isosceles.
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