Prove that this segment bisects another 
The circle touches the trapezoid $GFEC$ at the points $C$, $D$ and $E$. The point $A$ is the center of the circle. The rest of the information can be seen in the diagrams below. What we have to prove is that $FI=HI$.

I've added some diagrams below. The first one simply shows what is given and the others show what I've tried. I've tried a lot more than that but the other information  is basically useless (the information I've given here isn't quite useful either, though).


 
 A: Let $s$ the line defined by $FE$, $t$ the line defined by $CD$ And $J$ the point of intersection between $s$ and $t$. See the following figure:

Hints:
Note that $\angle GDC = \angle GCD = \angle FDJ = \angle DJF$.
Therefore $FJ=ED$.
But $FD=FE$ (Why?).
Note that $\triangle JFI \sim \triangle JEC$, hence ... 
A: Or let's coordinate spam this for the heck of it.
Define $A(0,0), E(0,r), C(0,-r),D(a,b)$ where $a<0$ and $b>0$ (as drawn)
Then we can find $\overline{FG}$ from its slope (as we know that the tangent line is perpendicular to the radius, so their slopes multiply up to -1) and from $D(a,b)$:
$\overline{FG}:(y-b)=-\frac{a}{b}(x-a)$
Solve for $F(F_x,r)$
$(r-b)=-\frac{a}{b}(F_x-a)$
$F_x=-\frac{b}{a}(r-b)+a$
And we can solve for $\overline{DC}$ because we have both C and D:
$\overline{DC}: y=\frac{b+r}{a}x-r$
Now solve for $I$ as the intersection of $\overline{DC}$ and $x=F_x=-\frac{b}{a}(r-b)+a$ to show that $I_y=0$
$I_y=\frac{b+r}{a}(-\frac{b}{a}(r-b)+a)-r=\frac{b^3-r^2b}{a^2}+b$
Then, substitute in $r^2=a^2+b^2$ gives you
$I_y=\frac{b^3-a^2b-b^3}{a^2}+b=-b+b=0$, for any choices of $a<0$ and $b>0$.
Now there's the answer, but guess that also proves that coordinate spamming takes way too long X_X
