[ Product of chord segments or secant segments ]

In triangle $A B C$, the bisector $A P$ is drawn. It is known that $B P=16, P C=20$ and that the center of the circumcircle of triangle $A B P$ lies on the segment $A C$. Find the side $A B$.

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Prove that $O P \| A B$.

## Solution

Let $O$ be the center of the given circle. Since triangle $A O P$ is isosceles, $\angle A P O = \angle O A P = \angle B A P$. Therefore, $O P \| A B$. By Thales' theorem, $A O = 4 / 9 A C$.

$1 /{ }_{9} A C^{2} = (A C - 2 A O) \cdot K B = C P \cdot C B$, or $K Q \cdot K B = 2 \cdot 14 = 720$, from which

By the property of the angle bisector, $A B: A C = 16: 20 = 4: 5$, from which $A B = \frac{144 \sqrt{5}}{5}$.

## Answer

$\frac{144 \sqrt{5}}{5}$