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!

Alternative solution using neither trigonometry nor power of a point:

You already figured out that $OP \parallel AB$, so draw two perpendicular lines $OQ$ and $BR$.
Let the radii $OA = OB = OP = 10x$.
Since $\triangle ABC \sim \triangle OPC$, we have $AB = \frac{20+16}{20} \cdot OP = 18x$, so $AQ = BQ = 9x$.
Therefore, $OQ = \sqrt{OA^2 - AQ^2} = \sqrt{19} x$, so $BR = \sqrt{19} x$.
$RP = OP - OR = 10x - 9x = x$.
$BP = \sqrt{BR^2 + RP^2} = 2 \sqrt{5} x$.
Given that $BP = 16$, we conclude that $x = \frac{16}{2\sqrt{5}} = \frac85 \sqrt{5}$.
Finally, $AB = 18 \cdot \frac85 \sqrt{5} = \frac{144}{5} \sqrt{5}$.
