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 $\tan\phi$. 
 A: Edited for revised question
Dropping the perpendicular from $C$ onto $AB$ will help.  Call the point $E$.  
Also drop the perpendicular from $P$ onto $BC$, and call the point $F$.  Then drop the perpendicular from $F$ onto $AB$, and call the point $G$.  
This gives a lot of similar and congruent triangles.

$$\tan \phi = \dfrac{|PF|}{|CF|} = \dfrac{|FB| }{ |CF|} = \dfrac{ |GB| }{|EG| } = \dfrac{ |PB| }{|AP| }= \dfrac{ c-|AP| }{|AP| }$$ so $$|AP| =  \dfrac{c}{ 1+\tan \phi}.$$
A: 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 $C=90^{\circ}\ \implies\ PA=\frac {c}{1+\tan\phi}$ .
A: 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.
