Hello, bearej50!
Given: Isosceles Triangle $\displaystyle ABC$ is inscribed in a circle with base $\displaystyle BC.$
Prove: If $\displaystyle P$ is any point on minor arc $\displaystyle BC$, then ray $\displaystyle PA$ bisects $\displaystyle \angle BPC.$ Code:
A
* o *
* / \ *
* / \ *
* / o \ *
/ \
* / \ *
* / o \ *
* / \ *
/ \
B*- - - - - -o- - -*C
* *
* *
* * * o
P
Draw chords $\displaystyle PB$ and $\displaystyle PC$.
Inscribed angle $\displaystyle APB$ is measured by $\displaystyle \tfrac{1}{2}\,\text{arc}(AB).$
Inscribed angle $\displaystyle APC$ is measured by $\displaystyle \tfrac{1}{2}\,\text{arc}(AC).$
Since $\displaystyle AB = AC$ .(the triangle is isosceles),
. . then $\displaystyle \text{arc}(AB) = \text{arc}(AC)$ .(equal chords subtend equal arcs).
Therefore: .$\displaystyle \angle APB = \angle APC \quad\Rightarrow\quad AP \text{ bisects }\angle BPC.$