Prove that in a cyclic quadrilateral $ABCP$ where two of its sides are equal the ratio $ {PA\over PB + PC}= {AC\over BC}$ is constant for any point P

Question

Let ABC be an isosceles triangle where $AB = AC$. Consider the circumference in which the points A, B and C live. Let P be a point in the chord of this circumference formed by the points B and C.

Prove that the ratio:

$$ \frac{PA}{PB + PC}= \frac{AC}{BC}$$

is constant for any point $P$.

Answer 1

From the Ptolemy's Theorem, $$(PA)(BC)=(PB)(AC)+(PC)(AB)$$ Since $ABC$ is iscoceles, $AB=AC$, so $$(PA)(BC)=(PB)(AC)+(PC)(AC)=(AC)(PB+PC)$$ And we get $$\frac{PA}{PB+PC}=\frac{AC}{BC}$$