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 $\tan\phi$. 

Interesting answers Henry and DonAntonio. Here comes a different approach from my friend(internet-friend) using trigonometry. It is nice actually.
Apply an well-known relation $\frac {PA}{PB}=\frac {CA}{CB}\cdot\frac {\sin\left(\widehat{PCA}\right)}{\sin\left(\widehat{PCB}\right)}=$ $\frac {\sin (C-\phi )}{\sin\phi}=$ $\frac {\sin C-\cos C\cdot\tan\phi}{\tan\phi}\implies$
$\frac {PA}{\sin C-\cos C\cdot\tan\phi}=\frac {PB}{\tan\phi}=\frac {c}{\sin C+(1-\cos C)\cdot\tan\phi}\implies$ $\boxed{PA=c\cdot\frac {\sin C-\cos C\cdot\tan\phi}{\sin C+(1-\cos C)\tan\phi}}$ .
Particular case $C=90^{\circ}\ \implies\ PA=\frac {c}{1+\tan\phi}$ .
