What's the measure of the segment $CG$ in question below?

Question

For reference: In a circumference of diameter $CE$,the strings $BE$ and $AC$ intersect at $F$ and the diameter $BD$ intercepts $AC$ in $G$. Calculate $GC$. If $AF = 3 , FG = 2$ and $AC = BE$

My progress: It seems to me little data... I don't know if it's possible to solve it with just this information

I find : T. Menelaus: $\triangle BOE-AC\\ EF.BG.OC=FB.GO.\underbrace{EC}_{2OC} \rightarrow EF.BE = 2FB.GP $

Answer 1

$ \small \triangle CBE \cong \triangle EAC$, as both are right triangles and one of their sides and their hypotenuse are equal. That leads to $ \small \triangle CFE$ being isosceles with $ \small CF = FE \implies BF = AF = 3$.

Apply Menelaus's theorem in $ \small \triangle CEF$ with traversal $ \small OB$ intersecting sides $ \small CE$ and $ \small CF$.

$ \displaystyle \small \frac{EB}{BF} \cdot \frac{FG}{GC} \cdot \frac{CO}{OE} = 1 ~ ​\implies \displaystyle \small \frac{EB}{3} \cdot \frac{2}{GC} = 1$

Using $ \small EB = AC, ~ 2 AC = 3 GC \implies 2 (GC + 5) = 3 GC$

$\therefore \small GC =10$

Answer 2

Consider the image with the fact that $\small \triangle CBE\cong\triangle EAC\ (\because AC=BE)$.

$\small{\triangle OCG\sim\triangle FBG}\implies\displaystyle{\frac3r=\frac2b}\implies 3b=2r$

From sine rule to $\small \triangle OCG$, $\dfrac{a}{\sin2x}=\dfrac b{\sin x}\implies a=2b\cos x\tag1$

From right triangle $\small CAE\to CA=CE\cos x\\$ $a+5=2r\cos x=3b\cos x\tag2$

$\frac{(1)}{(2)}$,$$\frac a{a+5}=\frac 23\implies \boxed{a=10}$$