The center of the circumcircle lies on a side of a triangle Consider a triangle $ABC$. Let the angle bisector of angle $A$ be $AP,P\in BC$. $BP=16,CP=20$ and the center of the circumcircle of $\triangle ABP$ lies on the segment $AC$. Find $AB$.

 $$AB=\dfrac{144\sqrt5}{5}$$


By Triangle-Angle-Bisector Theorem $$\dfrac{BP}{PC}=\dfrac{AB}{AC}=\dfrac{16}{20}=\dfrac{4}{5}\\ \Rightarrow AB=4x, AC=5x.$$ The cosine rule on $ABC$ gives $$BC^2=AB^2+AC^2-2\cdot AB\cdot AC\cdot\cos\alpha \\ \iff 1296=41x^2-40x^2\cos\alpha,$$ where $\measuredangle A=\alpha.$ Is any of this helpful for the solution? Any help would be appreciated. Thank you in advance!

Observe that $\angle OPA=\angle PAO=\frac\alpha2=\angle BAP\implies OP\parallel AB$. Thus $$\frac{CO}{OA}=\frac{CP}{PB}\iff \frac dR=\frac{20}{16}=\frac54$$ Also, Power of a point yields $$\begin{align*}\text{Pow}(C)_{(APB)}=\lvert d^2-R^2\rvert&=20\cdot 36\\\iff \left\lvert\left(\frac54R\right)^2-R^2\right\rvert&=720\\\iff \frac9{16}R^2&=720\\\iff R&= 16\sqrt{5}  \end{align*}$$ Use $OP\parallel AB$ again in order to infer

$$AB=R\cdot \frac{36}{20}=16\sqrt5\cdot \frac95=\frac{144\sqrt5}{5}$$

