# Find the ratio of EF/FC

Let E be the midpoint of side AB of square ABCD. Let the circle through B with center A and segment EC meet at F. what is the ratio of $CE/EF$?

Interestingly enough, it seems like setting a point G, where G is the midpoint of line BC, can form a line DG that intersects perpendicularly with line CE and intersects exactly at F. Should this be true, then similar triangle ratios can be used to determine the ratio CE/EF. How should I show that this is true? Or is there a better way to solve this problem?

• Hint: What is the angle $CEB?$
– user418131
Sep 18, 2018 at 14:38
• What if I suppose that calculators cannot be used, will finding angle $CEB$ still help? Sep 18, 2018 at 14:46
• Do you need one? If you have an expression for it, see where it leads you
– user418131
Sep 18, 2018 at 14:51
• Let me clarify my last comment - do you need the value of the angle? Will knowing it's trigonometric ratios do?
– user418131
Sep 18, 2018 at 14:59
• It seems like I was able to find an elementary solution involving transformation that completely avoids those, thank you for the comment anyways Sep 18, 2018 at 15:24

Let the side length of the square be $2$ and its vertex $A$ be at the origin:
$\hspace{2cm}$
The point $F$ is the intersection of the circle and the line: $$\begin{cases}x^2+y^2=4\\ y=2x-2\end{cases}\Rightarrow F\left(\frac85,\frac65\right).$$ Using the similarity of $\Delta BCE$ and $\Delta EFG$: $$\frac{CE}{EF}=\frac{CB}{FG}=\frac{2}{\frac65}=\frac53.$$
Let $BX$ be a diameter of the circle. We have that $BE:EX=1:3$. Using power of the point theorem, $EF\cdot EC=BE\cdot EX=\frac{3}{4}AB^2$. Now, $CE=\sqrt{5}AB/2$ and $EF=\frac{3}{2\sqrt{5}}AB$. Finally, $CE:EF=5:3$