Question

In triangle ABC, $\angle ABC={90}^{\circ }$ and P is a point on AC such that $\angle PCB=\angle PBC$, then prove that PA = PB.

Answer 1

Let angle $\angle PCB=\angle PBC=x$
In the right angled triangle,
$\angle ABC={90}^{\circ }$
$\angle ACB=x$

$\angle BAC={180}^{\circ }-({90}^{\circ }+x)$
$={90}^{\circ }-x$ .. (i)
And,
$\angle ABP=\angle ABC-\angle PBC$
$={90}^{\circ }-x$ ..( ii)
From (i) and (ii), we get
$\angle BAC=\angle ABP$
PA = PB
[Sides opposite to equal angles are always equal]