Perpendicular versus perpendicular bisector We have $AH=HB$ and $BG=GC$ in the image below. Why is $AD=2\times FG$?

 A: $\triangle BEA$ is similar to $\triangle BGK$ due to the right angle and the common angle at $B$. For the same reason, $\triangle BCI$ is similar to $\triangle BJH$. So the map characterized by
$$B\mapsto B\quad J\mapsto C\quad K\mapsto A\quad G\mapsto E\quad H\mapsto I\quad F\mapsto D$$
might be a single homothety. To make sure that it actually is, you'd have to verify that the scale factors for the two triangle similarities I mentioned actually agrees. Or equivalently that the points $B,D,F$ are on a single line.
When I wrote this answer, I thought that the scale factor was obviously $2$, but that was because I mixed up some points. I'll leave this as an incomplete answer for now since others refer to it.
A: It's not.  As MvG points out, you've got similar triangles, but the homothety he correctly identifies, does not have scale factor $2$. In order for the scale factor to be $2$, one would need $G \mapsto C$ and $H \mapsto A$.
As ratios:
$$\frac{AD}{KF} = \frac{AB}{KB} \ne \frac{AB}{HB} = \frac{2}{1}.$$
A: EDIT:  Answer updated to reflect the correction of the problem statement.
Draw a line through $G$ perpendicular to $AB$, and a line through $H$ perpendicular to $BC$.  Let $L$ be the intersection of these two lines.  You can then show that the figure $BHLG$ is proportional to $BADC$, and that the scale factor is $1:2$.  Therefore $AD = 2\times HL$.
But $HLGF$ is a parallelogram, and so $HL = FG$, from which your answer follows.
[Remnants of old answer below, because some of the comments make sense only in regard to this answer:]
The [previous] statement [that $AD = 2\times FG$] is not true in general.  For counterexamples, try letting $\triangle ABC$ be an equilateral triangle, or let $\angle ACB$ be a right angle.
There may be information missing from the problem statement, perhaps something to do with the circumscribed circle. The fact that there is a circumscribed circle tells us nothing--every triangle has one--so perhaps there is something else we are supposed to know about that circle?
A: Denote $AB=a$, $BC=b$, $\dfrac{AB}{BE}=k$.
Then
$BI = \dfrac{b}{k}$  $\implies$ $AI=AB-BI=a-\dfrac{b}{k}$;
$BJ = \dfrac{a}{2}\cdot k$ $\implies$ $GJ=BJ-GC=\dfrac{ak}{2}-\dfrac{b}{2}$.
$\triangle ADI \sim \triangle JFG$, so
$$
\dfrac{AD}{FG} = \dfrac{k\cdot DI}{FG} = k \cdot \dfrac{AI}{JG}=2k\times \dfrac{a-b/k}{ak-b} = 2. 
$$
