Isosceles triangle Let $ \triangle ABC $ be an $C$-isosceles and $ P\in (AB) $ be a point so that $ m\left(\widehat{PCB}\right)=\phi $. Express $AP$ in terms of $C$, $c$ and $\tan\phi$.
Edited problem statement(same as above but in different words):
Let $ \triangle ABC $ be a isosceles triangle with right angle at $C$. Denote $\left | AB \right |=c$. Point $P$ lies on $AB(P\neq A,B)$ and angle $\angle PCB=\phi$. Express $\left | AP \right |$ in terms of $c$ and $\tan\phi$. 

Putting $\,CE=$the height to the base in the triangle (and thus $\,CE\perp AB\,$) ,and putting $\,x:=\angle PCE\,\,,\,\,y:=\angle ECB\,$ , we get $\,\phi=x+y\,$ , and
$$\tan x= \frac{PE}{CE}\,\,,\,\tan y=\frac{c}{2\cdot CE}\Longrightarrow$$
$$AP=\frac{c}{2}-PE=\frac{c}{2}-CE\tan x=\frac{c}{2}-\frac{c}{2\tan y}\tan x\Longrightarrow$$
$$AP=\frac{c}{2}\frac{\tan y-\tan x}{\tan y}$$
But
$$\tan\phi=\tan(x+y)=\frac{\tan x+\tan y}{1-\tan x\tan y}$$
Try to take it from here.
