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).


 

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
