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$AB \perp CD$, $AE\perp CB$, $CE = BE$, $BE = a$, $FE = b$

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.

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up vote 2 down vote accepted

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 $\frac{a^2-b^2}{c}$.

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Thanks a lot, once again :) Also thanks to everybody else as well. – Lior Apr 14 '12 at 1:47

$$ AG = 2R = \frac{a^2+b^2}{b}$$

$$ FG=2b $$

$$\Rightarrow AF = \left(\frac{a^2+b^2}{b} -2b \right)= \frac{a^2-b^2}{b}$$

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 $c = \sqrt{a^2+b^2} = CF$

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$|\overline{CE}|=|\overline{BE}|=a$ 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, $|\overline{CH}|=2a\dfrac{a}{\sqrt{a^2+b^2}}$.

$|\overline{FH}|=|\overline{CH}|-|\overline{CF}|=\dfrac{2a^2}{\sqrt{a^2+b^2}}-\sqrt{a^2+b^2}=\dfrac{a^2-b^2}{\sqrt{a^2+b^2}}$

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