<s>[INST] 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!
 [/INST] $\angle POC = \angle A$, so $\triangle COP \sim \triangle CAB$
Hence $\frac{R}{AB} = \frac{20}{36} = \frac{5}{9}$ ...(i)
$\angle APB = \frac{1}{2} \angle AOB = 90^0 - \angle A$
$AB = 2R \sin \angle APB = 2R \cos A$
So from (i), $\frac{R}{2R \cos A} = \frac{5}{9}$
$\cos A = \frac{9}{10} = 1 - 2 \sin^2{\frac{A}{2}} \implies \sin \frac{A}{2} = \frac{1}{2\sqrt5} $
$16 = 2R \sin \frac{A}{2} \implies R = 16\sqrt5 $
$AB = 2R \cos A = 32\sqrt5 \times \frac{9}{10} = \frac{144}{\sqrt5}$
</s>