In triangle $ABC$, $AP$ is the angle bisector In triangle $ABC$, $AP$ is the angle bisector of $\measuredangle A$. If $BP=16,CP=20$ and the center of the circumcircle of $\triangle ABP$ lies on $AC$, find $AB$ and $AC$.

$AP$ is the angle bisector of $\measuredangle BAC$ which means that $\measuredangle BAP=\measuredangle CAP=\alpha$. On the other hand, $OA=OP=R_{\triangle ABP}=R$, so $\measuredangle PAO=\measuredangle APO=\alpha$. This makes $OP\parallel AB$. I am trying to figure out the best way to use that parallelity.
The most straightforward think we can note is $\triangle OPC\sim\triangle ABC$. I don't see if this can be helpful, though.
Another true thing is $\dfrac{BP}{PC}=\dfrac{AB}{AC}=\dfrac{16}{20}=\dfrac{4}{5}$.
 A: Let's say the radius of the circle is $r$. Call the intersection of $OC$ with the circle, $D\neq A$.
Using similarity, we have
$$\frac{r+DC}{r}=\frac{20}{16}$$
$$4(r+DC)=5r$$
$$DC=\frac{r}{4}$$
Evaluating the power of point $C$, we get that
$$CD\cdot AC=20\cdot 36$$
$$\frac{r}{4}\cdot \frac{9r}{4}=20\cdot 36$$
$$r=16\sqrt{5}$$
$$AC=\frac{9r}{4}=\boxed{36\sqrt{5}}$$
Using angle bisector theorem, we have
$$16\cdot AC=20\cdot AB$$
$$AB=\frac{4}{5}\cdot 36\sqrt{5}$$
$$AB=\boxed{\frac{144\sqrt{5}}{5}}$$
A: Let $AB =  4x, AC = 5x$ from the angle bisector theorem. Using the cosine rule on $\Delta ABC$, $36^2 = (4x)^2 + (5x)^2 - 2(4x)(5x) \cos 2a$ $ = x^2 (41 - 40 \cos 2a)$.
Now $\angle AOP = 180º - 2a$, so the reflex angle at the centre is twice the angle at the circumference, or $\angle ABP = \frac{1}{2}(360º - (180º - 2a)) = 90º + a$. Using the sine rule:
$$\frac{\sin(90º + a)}{5x} = \frac{\sin 2a}{36}$$
$$36 \cos a = 5x \cdot 2 \sin a \cos a, \ 18 = 5x \sin a.$$
Substituting this  value of $x$ gives:
$$36^2 = \frac{18^2}{25 \sin^2 a} (41 - 40 \cos 2a)$$
$$100 \sin^2 a = 41 - 40(2 \cos^2 a - 1)$$
$$100 - 100 \cos^2 a = 81 - 80 \cos^2 a$$
$$19 = 20 \cos^2 a$$
thus $\cos 2a = 2 \cos^2 a - 1 = \frac{9}{10}$ and $x^2 = \frac{36^2}{41 - 40 (9/10)} = \frac{36^2}{5}, x = \frac{36}{\sqrt5}$. Hence $AB = 4x = \frac{144}{\sqrt5}$, $AC = 5x = 36 \sqrt{5}$.
A: Yet an other solution using geometric constructions, similarities, and the theorem of Pythagoras...

Let $OP=OA=OB$ be $20t$, with some unknown $t$.
Then $AB:20t=AB:OP=CB:CP=36:20$, so $AB=36t$.
Let $R$ be the mid point of $AB$. Then $AR=RB=18t$. In $\Delta OAR$ there is a right angle in $R$, so $OR=2t(10^2-9^2)=2t\sqrt{19}$.
Let now $C'$ be the projection of $C$ on $AB$. From $AC':18t=AC':AR=AC:AO=9:4$ we obtain $AC'=\frac12\cdot 81t$. So $BC'=AC'-AB=\frac 12 \cdot 9t$.
Finally, $CC:2t\sqrt 19=CC':OR=AC:AO=9:4$ gives $CC'=\frac 12\cdot 9t\sqrt {19}$.
All this data is displayed in the following picture:

In $\Delta CBC'$ we can now determine $t$ from
$$
36^2 = BC^2 = C'B^2+C'C^2=\frac{9^2t^2}4(1+19)=9t^2\cdot 5\ .
$$
This gives $t=\frac 4{\sqrt 5}=\frac 15\cdot 4\sqrt 5$, so:

*

*$\color{blue}{AC=45t=36\sqrt 5}\approx 80.4984471\dots$ , and

*$\color{blue}{AB=36t=\frac 15\cdot 144\sqrt 5}\approx 64.398757751\dots$ .


Note: A posteriori we have the information that the involved quantities would have been slightly more beautiful when starting with $40s$ instead of $20t$, but i decided to keep the initial $20t$, since the denominator $2$ cannot be anticipated with bare eyes...
A: Here is a solution without using power of point explicitly.
Let radius of circle be $r$, $AC=5k$ and $AB=4k$. Then, using $\Delta COP \sim \Delta CAB$, we get-
$$\frac {r}{4k}=\frac {20}{36}$$
Which means that $$k=\frac {9r}{20} {\tag 1}$$
Also, from $\Delta AOP$, we get: $AP=2r \cos {\frac A2}$.
Now applying sine law on $\Delta APB$,
$$\frac {2r \cos \frac A2}{\sin B}=\frac {16}{\sin \frac A2} {\tag 2}$$
Also, note that angle subtended by a chord at centre is double the angle subtended at circumference. Hence if you consider the chord $AP$ and $\angle PBA$, we get that $\angle AOP=2\pi-2B$. This means that for $\Delta AOP$, we have:
$$\angle \frac A2+2\pi-2 \angle B +\angle \frac A2=\pi$$
Which means that $B=\frac {\pi}{2}+\frac A2$.
Substituting this in $(2)$ simplifies to:
$$r=\frac {8}{\sin \frac A2} {\tag 3}$$
Join $OB$.Then for $\Delta OAB$, we have, from trigonometry and $(1)$:
$$AB=2r\cos A=\frac {9r}{5}$$
Which means that $\cos A=\frac  {9}{10}$, so $\sin \frac A2=\frac {1}{2\sqrt 5}$.
Substituting in $(3)$, we get:
$r=16 \sqrt 5$.
This means that $AC=\frac {9r}{4}=36 \sqrt 5$ and $AB=\frac {9r}{5}=\frac {144}{\sqrt 5}$.

Perhaps this answer well demonstrates the importance of the power-of-point technique. Note the variety of techniques including trigonometry, sine law, and properties of cyclic quadrilaterals, I had to use to bypass the power-of-point.
A: Let the radius of the circle be $R$, and let $AC = 2R + x$. Now by similarity you have $\displaystyle \dfrac{R+x}{2R+x} = \dfrac{CP}{CB}$, and by the secant theorem, $x(2R+x) = CP\cdot CB$. The rest is pure algebra.
