how to prove the segment $FH=HE$ 
Given a right-triangle $\triangle AGC$ $(\angle AGC=90^\circ)$ point $D$ is an arbitrary point on the altitude. $AE=AG$ and $CF=CG$.
How to prove that $FH=HE$?

I found that circles centered at $A$ and $C$ are orthogonal circles. The altitude of triangle $\triangle AGC$ is the radical axis of the 2 circles and the circle of radius $FH$ is tangent to both (green) circles, but I couldn't prove $FH=HE$.
Please help.

 A: 
LEMMA: Extend segment $ED$ until it meets circle with center $A$ and radius $AG$ at point $J$. Draw median $AB'$ of chord $EJ$ until it intersects the radical axis $GK$ at point $B$. Point $B$ is the pole for polar line $EJ$.
PROOF: It's easy to see that:
$$\triangle AB'C\sim \triangle AKB,\quad \triangle AKG\sim AGC$$
It means that:
$$\frac{AB'}{AC}=\frac{AK}{AB},\quad \frac{AK}{AG}=\frac{AG}{AC}$$
$$AB'\cdot AB=AK\cdot AC=AG^2$$
It follows that point $B$ is the inverse point of point $B'$ with respect to circle with center $A$ and radius $AG$. By definition point $B$ is the pole for polar line $EJ$. As a consequence, lines $BE$ and $BJ$ (not shown) are tangent to the circle with center $A$.
End of lemma proof

Back to the main problem: Construct point $B$ as explained in the lemma. According to lemma proof line $BE$ is tangent to circle with center $A$ and radius $AG$.
Let's do a similar thing for the other circle with center $C$. First, extend line $FD$ until it interesects circle with center $C$ and radius $CG$ at point $L$. Then draw segment $BC$.
Note that point $D$ is the orthocenter of triangle $ABC$ ($BK\perp AC$, $CB'\perp AB$). It follows that $AL\perp BC$. In other words, line $CB$ is the median of chord $FL$.
According to our little lemma it means that point $B$ is also the pole for polar line $FL$. Because of that, line $BF$ is tangent to the circle with center $C$ and radius $CG$.
The rest is trivial. Triangles $BEH$ and $BFH$ are congruent by SSA: $BH=BH$, $BE=BF$ (because point $B$ is on the radical axis of the two circles) and $\angle BEH=\angle BFH=90^\circ$. Therefore:
$$HE=HF$$
Very nice problem, BTW.
