Here in the figure, we have:
$\bullet$ The radii of both the circles are equal
$\bullet$ $E$ bisects both $\overline{AC}$ and $\overline{FG}$
$\bullet$ $\overline{HI}$ is a line passing through $E$
We have to prove that $E$ bisects $\overline{HI}$ as well.
(I think it is really obvious from the figure that's the case but I'm having a hard time proving it.)
My attempt:
My first attempt was to try to prove that $\triangle AEI\cong\triangle CEH$
I found,
in $\triangle AEI$ and $\triangle CEH$
$\overline{AE}=\overline{EC}$ [$\because$ $E$ bisects $\overline{AC}$]
$\angle AEI=\angle HEC$ [$\because$ vertically opposite angles]
$\overline{AI}=\overline{HC}$ [$\because$ radii of the circles are equal]
but because these three conditions don't correspond to any congruence criterion, I'm unable to proceed.
I also tried to prove $\angle HCE=\angle EAI$ and hence $\overline{AI}\parallel \overline{HC}$ but failed.
Then I made some constructions which are shown here.
I drew $\angle H'AE=\angle IAE$ and $\angle HCE=\angle I'CE$
That way I was able to prove $\triangle AH'E\cong\triangle AIE$ and $\triangle HCE\cong\triangle I'CE$
so,
$\overline{H'E}=\overline{IE}$
and
$\overline{HE}=\overline{I'E}$
but again, I'm unable to show that the two pairs are equal. I'm unable to proceed any further.
I'd be very glad if someone could help me prove it.
Thanks a lot!
Reflect $H$ along $E$ to get the point $M$. Consider $\triangle AEM$ and $\triangle CEH$, we have, $$AE=CE\ \ \ (given)\\ EM=EH\ \ \ (by\ \ construction)\\ \angle AEM=\angle CEH\ \ \ (vertically\ \ opposite\ \ angles)$$ Hence, by $SAS$ congruence criterion, we have, $$\triangle AEM\cong\triangle CEH$$ Therefore, $$AM=CH$$ Now, since the radii of both the triangles are equal, this means that $M$ lies on the circumference of the left circle, therefore, it is coincident with $I$.
Hence proved