Points $A$, $B$, $C$, $D$ lie on a circle (in that order) where $AB$ and $CD$ are not parallel. The length of arc $AB$ (which contains the points $D$ and $C$) is twice the length of arc $CD$ (which does not contain the points $A$ and $B$). Let $E$ be a point where $AC=AE$ and $BD=BE$. Prove that if the perpendicular line from point $E$ to the line $AB$ passes through the center of the arc $CD$ (which does not contain the points $A$ and $B$), then $\angle ACB = 108^\circ$.
Shu wrote:
The length of arc $AB$ (which contains the points $D$ and $C$) is more than twice the length of arc $CD$ (which does not contain the points $A$ and $B$).
I think it has to be exactly twice.
Let $M$ be the midpoint of arc $CD$. Then $AM^2-MB^2=AE^2-EB^2=AC^2-CB^2$ so $CA^2-AM^2=DB^2-BM^2$. Let $F$ be the reflection of $B$ about the perpendicular bisector of $CD$. Then $F\equiv A$. We also have $CF^2-FM^2=CA^2-AM^2$, so $FA\perp CM$. Let $\angle DCM=x$. Then $\angle ACB=\angle DCB-\angle DCA=\angle CDF-\angle ACD=\angle FMC-\angle ACD=90+\angle AFM-\angle ACD=90+x$. But also $\angle ACB=180-4x$, so $x=18$ and $\angle ACB=108$.
Thank you and I apologize for the error in translation. (I was wondering why I had such a hard time drawing an accurate diagram. ) It's been corrected.