Let $ABCD$ be a square.
$E$ is a point outside $ABCD$.
Connect $AE$ satisfying $AC=AE$.
$F$ lies on $AE$ and $DE=DF$.
Connect $BF$ which intersects $AC$ at $G$ and $BG=EC$.
Show that $BG\parallel EC$.
My Attempt
- Suppose that the square has side length $s$.
- The circle $O(A, AC)$ with center at $A$ and radius $AC$ is $x^2+(y-s)^2=(\sqrt{2}s)^2$.
- Since $E$ is on the circle $O(A, AC)$, we can assume $E=\left(p, -\sqrt{2s^2-p^2}+s\right)$.
- The circle $O(D, DE)$ with center at $D$ and radius $DE$ is $(x-s)^2+(y-s)^2=(p-s)^2+\left(-\sqrt{2s^2-p^2}\right)^2$.
- The line $AE$ is $\frac{y-s}{x}=\frac{-\sqrt{2s^2-p^2}}{p}$.
- The intersections of $O(D, DE)$ and $AE$ are $E$ and $F=\left(\frac{p^2-ps}{s}, \frac{-(p-s)\sqrt{2s^2-p^2}+s^2}{s}\right)$.
- The line $AC$ is $y=-x+s$.
- The line $BF$ is $\frac{y}{x}=\frac{-(p-s)\sqrt{2s^2-p^2}+s^2}{p^2-ps}$.
- The intersection of $AC$ and $BF$ is $G=\left(\frac{sp^2-ps^2}{-(p-s)\sqrt{2s^2-p^2}+s^2+p^2-ps}, \frac{-s(p-s)\sqrt{2s^2-p^2}+s^3}{-(p-s)\sqrt{2s^2-p^2}+s^2+p^2-ps}\right)$.
- Since $BG=CE$, we have...the equation is too complicated to solve.
Assuming that $F$ lies in the interior of the triangle $ACD$, the assertion may be proven in two steps.
Uniqueness: Extend $AD$ to intersect circle $(A,AC)$ at $H$ on the right. Let $E$ lie on arc $CH$ so $AC=AE$ always holds. Now move $E$ along arc $CH$. If $E$ is close to $C$, then $G$ lies outside of circle $(B,CE)$, but if $E$ is close to $H$, then $G$ lies inside of circle $(B,CE)$. By continuity, there is an occasion such that $G$ lies on circle $(B,CE)$ so that $BG=CE$. It may be tedious to show that such point is unique, but the idea is clear (so details are omitted).
Existence: It suffices to require that $FBC$ is an equilateral triangle (or equivalently $\angle EAC=30^\circ$). Then by angle chasing, it is easy to see that $ABF,AGF,ACE$ are isosceles triangles with base angle equal to $75^\circ$. Since $$\triangle AGF\sim \triangle ACE$$ one has $GF\parallel CE$, so $BG\parallel CE$, as required. QED