Question

In $\Delta ABC,\angle$ $ABC={90}^{\circ }$ and P is a point on AC such that $\angle PBC=\angle PCB$. Show that : $PA=PB.$ Enter $1$ if true else $0$

Answer 1

In $△ABC$,
$\angle A+\angle B+\angle C=180$
$\angle A+90+\angle C=180$
$\angle A=90-\angle C$
Also, $\angle ABP=\angle B-\angle PBC$
$\angle ABP=90-\angle PBC$
$\angle ABP=90-\angle C$ (Given PB = PC, by Isosceles triangle property, $\angle PBC=\angle C$)
Hence, $\angle ABP=\angle A$
or, $PA=PB$