Trig‎o‎‎‎‎nometry‎‎‎‎‎‎‎ Question‎‎‎‎‎‎‎ $AB \perp CD$, $AE\perp CB$, $CE = BE$, $BE = a$, $FE = b$
In the first few parts of the question I've answered the following:
$AF = AD$
$FE = GE$
$R = \frac{a^2+b^2}{2b}$
The last part of the question, the one I'm having trouble solving, is finding $FH$.

Thanks.
 A: We chase distances a little, and the answer pops out. Note that $\triangle CEF$ and $\triangle CHB$ are similar.  Calculate everything, well, everything that matters.
We have $CF=\sqrt{a^2+b^2}$. Out of sentiment call that $c$. We have $\frac{CH}{CB}=\frac{CE}{CF}=\frac{a}{c}$. Since $CB=2a$ we have $CH=\frac{2a^2}{c}$.  To find $FH$, subtract $c$. Now it's over, $FH=\frac{2a^2}{c}-c$. 
One might wish to simplify this to $\frac{2a^2-c^2}{c}$, and then to $\frac{a^2-b^2}{c}$.
A: $$ AG = 2R = \frac{a^2+b^2}{b}$$
$$  FG=2b $$
$$\Rightarrow AF = \left(\frac{a^2+b^2}{b} -2b \right)= \frac{a^2-b^2}{b}$$
By Intersecting Chords Theorem
$$AF . FG = CF . FD $$
$$FD = 2FH$$
Therefore 
$$FH = \frac{ \left(\frac{a^2-b^2}{b}\right) 2b}{c} = \frac{a^2-b^2}{c}$$
where $c = \sqrt{a^2+b^2} = CF$ 
A: $|\overline{CE}|=|\overline{BE}|=a$ and $|\overline{FE}|=b$, so Pythagoras says $|\overline{CF}|=\sqrt{a^2+b^2}$. Furthermore, $|\overline{CB}|=|\overline{BE}|+|\overline{CE}|=2a$.
Similar triangles yield $\frac{|\overline{CH}|}{|\overline{CB}|}=\frac{|\overline{CE}|}{|\overline{CF}|}$; therefore, $|\overline{CH}|=2a\dfrac{a}{\sqrt{a^2+b^2}}$.
$|\overline{FH}|=|\overline{CH}|-|\overline{CF}|=\dfrac{2a^2}{\sqrt{a^2+b^2}}-\sqrt{a^2+b^2}=\dfrac{a^2-b^2}{\sqrt{a^2+b^2}}$
