For reference: In the isosceles triangle $ABC$ of incenter $I$, on the arc IC of the circle circumscribed to AIC, the point P is point P so that $AP = 10$ and $PC= 6$; calculate AB (AC is the base of the isosceles triangle ABC and $B=74^o$)
My progress:
Apollonius's theorem:
$\triangle APC: 6^2+10^2 = 2PM^2+\frac{B^2}{2}\rightarrow 272 = 2PM^2+B^2\\ \triangle IMC: IM^2+(\frac{b}{2})^2 = IC^2\\ \triangle ABC:\frac{BI}{IM}=\frac{a+c}{b}\\ \triangle ABM: c^2 = BM^2 + (\frac{b}{2})^2\\ 10.6=2R.IM\implies R.IM = 30\\ \triangle AIM:notable(\frac{53^o}{2})\implies \frac{b}{2}=2IM\therefore b = 4IM\\$
...?
You can simply apply law of cosine in $\triangle APC$ to find $AC$. Also knowing that the side lengths are approximately in ratio $3:4:5$ in a right triangle with angles $37^\circ$ and $53^\circ, ~$ you can drop a perp from $A$ to $CP$ extend.
$\angle APC = \angle AIC = 127^\circ \implies \angle APX = 53^\circ$. That leads to $AX \approx 8, PX \approx 6$
$AC^2 = AX^2 + CX^2 \approx 8^2 + 12^2 \implies AC \approx 4 \sqrt{13}$
So, $AM \approx 2 \sqrt{13}$ and $AB \approx \frac53 \cdot AM \implies AB \approx \frac{10 \sqrt{13}}{3}$