Prove that $FG$ is tangent to a fixed circle.

Question

Let $ABCD$ be a rectangle with $2AB=BC$.

$E$ is a mobile point on $AD$.

$AB\cap CE=\{G\}$ and $CD\cap BE=\{F\}$.

Prove that $FG$ is tangent to a fixed circle.

Can someone help me? I don't know how to start this type of problem.

Answer 1

Say that $DF=ax$, then due to similarity, $AG=\frac{x}{a}$. Then we calculate $FG=\sqrt{(2x)^{2}+(ax-\frac{x}{a})^{2}}=ax+\frac{x}{a}$

We now know that $BG+FC=BC+FG$ and therefore $BCFG$ is a quadrilateral circumscribing a circle. Furthermore, $BG//CF$ with distance $2x$, thus the circle is always the same circle with $AD$ as diameter regardless of position of $E$