\(P\) 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.
Proof:
Given that \(P\) is a point on the bisector of \(\angle ABC\). The line through \(P\), parallel to \(BA\) meet \(BC\) at \(Q\).
\(\angle ABP = \angle PBC\)
That is, \(\angle 1 = \angle 2\) ---- (\(1\))
Also, from the figure, we observe that \(PQ \parallel AB\) and \(PB\) is the transversal line. Then, we have:
\(\angle ABP = \angle \)
That is, \(\angle 1 = \angle 3\) ---- (\(2\))
From equations (\(1\)) and (\(2\)), we have:
\(\angle 2 = \angle 3\)
That is, \(\angle PBC = \angle BPQ\)
We know that, sides opposite to angles are equal
Thus, \(BQ = \)
Therefore, \(BPQ\) is an isosceles triangle.
Hence, we proved.