Geometry: proving $B$, $K$, $D$, $A$ concyclic

Question

Let $AC$ be the diameter of $\omega$ which passes through $D$. The perpendicular bisector of $CD$ intersects $\omega$ at $F$, $G$. A line through $G$ intersects $FD$ at $K$ and $\omega$ at $B\ne G$. Show that $B$, $K$, $D$, $A$ are concyclic.

I first tried angle chasing, which means proving $\angle B=\angle FDE$ etc. but with no success.

Then I thought of the method of moving points. We move point $K$. The transformation $K\to B$ preserves cross ratio, which means that it suffices to check $3$ special cases. But I only managed to check $2$.

Answer 1

You found $\widehat {B}=\angle FDE$ is sufficient to conclude points B, K, D and A are con-cyclic, because we have:

$\angle FED+\angle FDA=\angle KBA+\angle FDA=180^o$

This is due to this fact that every quadrilateral the sum of it's opposite angles is $180^o$ is con-cyclic.

As can be seen in figure:

$FD=DG=GC=CF\Rightarrow CG||FI$

$\Rightarrow \overset{\large \frown}{CF}=\overset{\large \frown}{IG}$

$\angle FDC=\frac{\overset{\large \frown}{CF}+\overset{\large \frown}{AI}}{2}=\frac{\overset{\large \frown}{GI}+\overset{\large \frown}{AI}}2=\angle GBA$