Question

$ABC$ is an isosceles triangle in which $AB=AC$. $P$ is any point in the interior of $\Delta ABC$ such that $\angle ABP=\angle ACP$. Prove that
a) $BP=CP$
b) $AP$ bisects $\angle BAC$

Answer 1

Given ABC is a trianlge in which $AB=AC$ . P is any point in the interior of the triangle such that angle $ABP=ACP$

In $\Delta APB$ and $\Delta APC$,

$AB=AC$[given]

$\angle ABP=\angle ACP$ [given]

$AP=AP$[common]

$\Delta APB\cong \Delta APC$[by SAS congruency criterion]

$\therefore \angle PAB=\angle PAC$ [corresponding angles of congruent trianlges]

thus, BP=CP

And AP bisects $\angle BAC$