To solve the given problem:

We need to prove that: $\frac{PA}{PB + PC} = \frac{AC}{BC}.$

Proof:

1. Given:

   • $\triangle ABC$ is isosceles, where $AB = AC$.
   • The triangle is inscribed in a circle, and $P$ lies on the arc $BC$ opposite $A$.

2. Using properties of a circle:

   • Since $\triangle ABC$ is inscribed, the angles subtended by the same arc are equal. This implies $\angle BPC = \angle BAC$.
   • Let $\angle BAC = \alpha$. Therefore, $\angle BPC = \alpha$.

3. Relations involving arcs:

   • $PA$, $PB$, and $PC$ are chords of the circle. Using the property of angles subtended by a chord in a circle, we have: $\frac{PA}{\sin \angle BPC} = \frac{PB}{\sin \angle APC} = \frac{PC}{\sin \angle APB}.$

4. Expressing $\angle BPC$:

   • As stated, $\angle BPC = \alpha$.

5. **Expressing ( PB + PC: \