Prove that length of two line segments are equal in a circle 
In the figure, B is the center of the semi-circle, $C$, $E$ are two points on the circle, $CG$ and $EF$ is perpendicular to $AF$, $ED$ is perpendicular to $CB$, prove that $DF$ = $CG$.

How to do this question without coordinate-geometry? Thanks in advance. Hints are welcomed.

Seems that this is useful, but I don't see how:

 A: Let $R$ be the radius of the semicircle.
In right triangle $CGB$, $ \frac{ CG} { BG} = \sin \angle CBG$.
EDBF is a cyclic quad, so by extended sine rule, $\frac{ DF}{ \sin \angle DBF } = R$.
Thus, $ DF = R \sin \angle DBF = R \sin \angle CBG = CG$.
A: Trigonometric Solution
$\sin \angle COD=\frac {CD}{R}$
$\sin \angle COD=\sin \angle (180-COD)=\sin \angle COF$
${\frac{GF}{\sin \angle COF}=\frac{OF}{\sin \angle COF}}$
$\angle FGO=\sin \angle OEF$ (same segments)
$\sin \angle OEF=\sin \angle FGO=\frac {OF}{R}$
${\frac{GF}{\sin \angle COF}=\frac{OF}{\frac{OF}{R}}}=R$
$\sin \angle COF = \frac {GF}{R}$
$\sin \angle COD=\frac {CD}{R}  = \frac {GF}{R}  =\sin \angle COF$
$CD=GF$
A: I wrote this in case anyone is interested in a proof that doesn't use the cyclic quad machinery.
One can see that $ \angle FEG = \angle COA $ since one is a $ 90^o $ rotation of the other.  Let $ \alpha = \angle FEG = \angle COA $ and $ \beta = \angle AOE $.  One can deduce (by applying the definition of sine and cosine to triangle $ OGE $) that $ |OG| = r\cos(\beta - \alpha) $ and $ |GF| = r\sin(\beta - \alpha) $, where $ r $ is the radius of the circle.
Now consider triangle $ FEG $.  Applying the law of cosines to this triangle, we have
$$
|EG|^2 = |FG|^2 + |FE|^2 - 2 |FG||FE|\cos(\angle EFG)
$$
Now, by direct computation we find that
$$
|EG|^2 = r^2 \sin^2(\alpha)
$$
which is the same as $ |CD|^2 $.
