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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 $\tan\phi$.

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What does $\,C-$isosceles, $\,(AB)\,$ and $\,m(\widehat{PCB})\,$ mean? That C is the base, (AB) one of the equal sides and the angle...or what? – DonAntonio Dec 29 '12 at 23:49
@DonAntonio I guess that $C$ is the vertex opposite to the base; the base is denoted by $(AB)$ and $m(\widehat{PCB})$ is the measure of the angle with vertex $C$. $c$ is the length of the base $(AB)$. – Sigur Dec 30 '12 at 0:20
That seems plausible, @Sigur, yet it is interesting the OP didn't bother to address the question... – DonAntonio Dec 30 '12 at 0:24
I see that I have confused many of you with my problem statement. I will edit it so it will be easier to understand. – EricAm Dec 30 '12 at 2:14
You wrote in your edition "...with right angle at $\,C\,$" this correct? Is it then a right angle isosceles triangle? – DonAntonio Dec 30 '12 at 4:45
up vote 1 down vote accepted

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.

enter image description here

$$\tan \phi = \dfrac{|PF|}{|CF|} = \dfrac{|FB| }{ |CF|} = \dfrac{ |GB| }{|EG| } = \dfrac{ |PB| }{|AP| }= \dfrac{ c-|AP| }{|AP| }$$ so $$|AP| = \dfrac{c}{ 1+\tan \phi}.$$

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Thanks for the answer Henry but unfortunately you have misunderstood the problem. I will edit my problem statement and hopefully it will become clearer. – EricAm Dec 30 '12 at 2:15

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 $C=90^{\circ}\ \implies\ PA=\frac {c}{1+\tan\phi}$ .

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So let's figure out why this and Henry's solutions differs. – EricAm Dec 30 '12 at 18:57
It seems one of my equalities was upside down – Henry Dec 30 '12 at 21:46

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

$$\tan x= \frac{PE}{CE}\,\,,\,\tan y=\frac{c}{2\cdot CE}\Longrightarrow$$

$$AP=\frac{c}{2}-PE=\frac{c}{2}-CE\tan x=\frac{c}{2}-\frac{c}{2\tan y}\tan x\Longrightarrow$$

$$AP=\frac{c}{2}\frac{\tan y-\tan x}{\tan y}$$


$$\tan\phi=\tan(x+y)=\frac{\tan x+\tan y}{1-\tan x\tan y}$$

Try to take it from here.

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