What's the ratio between the segments $\frac{AF.BG}{FG}$ in the figure below?

Question

For reference: In the figure below the trapezoid has height $13$ and is inscribed in a circle of radius $15$. Point $E$ is on the minor arc determined by $A$ and $B$, the points $F$ and $G$ intersect with $ED$, $E$C and $AB$. Then the ratio between the segments $\frac{AF \times BG}{FG}$ is?

My progress: I'm lost in this question...Perhaps using some of the segment $AB$ ratio with the diameter $DC$ ??

$\triangle EFG\sim HJC: \frac{JC}{FG} = \frac{JH}{FE}=\frac{HC}{EG}\\ \triangle BJG \sim \triangle HJC: \frac{JC}{JG}=\frac{JH}{BJ}=\frac{HC}{BG}\\ \triangle KID \sim \triangle KAF: \frac{JC}{FK}=\frac{JH}{AF}=\frac{HC}{AK}$

...???

Answer 1

Can you see that $~\triangle AFE \sim \triangle CBE~$ and $~\triangle BGE \sim \triangle DAE~$?

That leads to $~ \displaystyle \frac{BC}{AF} = \frac{CE}{AE}~, \frac{AD}{BG} = \frac{AE}{GE}$

Multiplying, $\displaystyle \frac{BC \times AD}{AF \times BG} = \frac{CE}{GE} \tag1$

As $ \displaystyle \triangle FEG \sim \triangle DEC, ~\frac{CE}{GE} = \frac{CD}{FG}$

Plugging into $(1)$,

$\displaystyle \frac{AF \times BG}{FG} = \frac{BC \times AD}{CD}$

As it is a cyclic trapezium, it must be isosceles and $BC = AD$. Applying Pythagoras,

$OH^2 = OB^2 - BH^2 = 15^2 - 13^2 = 56$

So, $CH = 15 - \sqrt{56}$

$BC \times AD = BC^2 = BH^2 + CH^2 = 13^2 + (15 - \sqrt{56})^2 = 30 (15 - 2 \sqrt{14})$

So the answer is $~(15 - 2 \sqrt{14})$

Answer 2

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Notice that points $H$ and $I$ in this answer are different from those in the original post.

Let $\frac{AF\cdot BG}{FG}=R$. First of all, $$AB=AF+FG+BG=2\sqrt{15^2-13^2}=4\sqrt{14}$$

Power of a point gives us \begin{align*} AF\cdot FB&=AF\cdot\left(4\sqrt{14}-AF\right)=EF\cdot DF&&(1)\\ BG\cdot AG&=BG\cdot\left(4\sqrt{14}-BG\right)=EG\cdot GC&&(2) \end{align*} and it follows that \begin{align*} \frac{(1)\cdot(2)}{FG^2}\implies R\left(4\sqrt{14}+R\right)=\frac{EF}{FG}\cdot \frac{EG}{FG}\cdot DF\cdot GC \end{align*}

Now, with $\triangle EFG\sim\triangle HDF$ we have $\frac{EF}{FG}=\frac{HD}{DF}$. Similarly, with $\triangle EFG\sim\triangle IGC$ we have $\frac{EG}{FG}=\frac{IC}{GC}$. Therefore, $$R\left(4\sqrt{14}+R\right)=HD\cdot IC$$

Since $\triangle HDF\sim\triangle IGC$, it follows that $$HD\cdot IC=HF\cdot IG=13^2=169$$ and we finally have the quadratic equation $$R^2+4\sqrt{14}R-169=0$$ whose solutions are $$R=-2\sqrt{14}\pm15$$ We'll take $\color{blue}{R=-2\sqrt{14}+15}$ since $R>0$.

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Maybe there are solutions that are more straightforward or concise, but I'm just gonna put my answer out here. I believe $R=-2\sqrt{14}+15$ is the correct answer since I've verified it on GeoGebra. Some lengthy brute-force simplifications are skipped in this answer.