If I have the following triangle:
Where $\angle B=\angle C = O$
And $AP$ bisects $\angle A$ so essentially $\angle BAP = \angle CAP = \frac12 \angle A$
We can prove that $\angle APB = \angle APC$ but how do we show that they are 90 degrees?
$$ \angle APB + \frac12 \angle A + \angle ABC = 180^\circ = \angle ABC + \frac12 \angle A + \angle APC $$
So:
\begin{align} \angle ABP &= 180^\circ - \left( \angle ABC + \frac12 \angle A \right) \\ \angle APC &= 180^\circ - \left( \angle ABC + \frac12 \angle A \right) \\ \end{align}
So $\angle ABP = \angle APC$ but is it also proven that they are 90 degrees? How?
First we write a formula for the sum of the angle measures of triangle PAB. I'm going to just use "APB" as "angle APB". We note that angle ABP is "O": $$O + \frac{A}{2} + APB = 180$$ Next we write a formula for the sum of the angle measures of triangle ABC: $$2\cdot O + A = 180$$ This gives you that $$O = \frac{180-A}{2}.$$ We substitue this value for "O" in the formula for triangle PAB: $$\frac{180-A}{2}+\frac{A}{2} + APB = 180$$ $$90 - \frac{A}{2}+\frac{A}{2}+APB = 180$$ $$90 + APB = 180$$ $$APB = 90.$$
We can do the same thing for APC, using triangle PAC in place of triangle PAB.