0
$\begingroup$

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?

$\endgroup$
5
  • $\begingroup$ Hint: What is the angle $CEB?$ $\endgroup$
    – user418131
    Sep 18, 2018 at 14:38
  • $\begingroup$ What if I suppose that calculators cannot be used, will finding angle $CEB$ still help? $\endgroup$ Sep 18, 2018 at 14:46
  • $\begingroup$ Do you need one? If you have an expression for it, see where it leads you $\endgroup$
    – user418131
    Sep 18, 2018 at 14:51
  • $\begingroup$ Let me clarify my last comment - do you need the value of the angle? Will knowing it's trigonometric ratios do? $\endgroup$
    – user418131
    Sep 18, 2018 at 14:59
  • $\begingroup$ It seems like I was able to find an elementary solution involving transformation that completely avoids those, thank you for the comment anyways $\endgroup$ Sep 18, 2018 at 15:24

2 Answers 2

1
$\begingroup$

Let the side length of the square be $2$ and its vertex $A$ be at the origin:

$\hspace{2cm}$enter image description here

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 $\Delta EFG$: $$\frac{CE}{EF}=\frac{CB}{FG}=\frac{2}{\frac65}=\frac53.$$

$\endgroup$
0
$\begingroup$

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, $CE:EF=5:3$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.