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

consider the figure above.

P is the point on the bisector of angle ABC. 

 ∠1 =∠ 2  [ BP is the bisector of ∠ABC]

Also  PQ is parallel to AB

∠1 = ∠3  [ If a transversal intersects two parallel lines, then alternate interior angles are equal ]

∴∠ 2 = ∠3

And since in a triangle if two angles are equal then its sides opposite to these angles are also equal.

 ∴BQ = PQ

Now, two sides of the ?BPQ are equal.

∴ΔBPQ is isosceles.

Hence Proved

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