Prove that $BF$ bisects the line segment $DE$ Let $C$ be a point on a semicircle $\Gamma$ of diameter $AB$ and let $D$ be the midpoint of the arc $AC$. Let $E$ be the projection of $D$ onto the line $BC$ and $F$ the intersection of the line $AE$ with the semicircle. Prove that $BF$ bisects the line segment $DE$

Here I created a point $G$ such that $GD$ is parallel to $EB$ so that I can change the problem to proving that $GF=FE$. I feel that this can help because $BF$ is normal to FG so we might use some radical axis. The problem here is that I don't know where to draw the other circle, maybe a circle with diameter $FB$? Also, I don't know how to use the condition $D$ is the arc $AC$.
 A: You don't need $G$. What you only need to do is that $\triangle DFB$ and $\triangle EFB$ have equal area. That is,
$$\frac 12DF\times FB\times \sin\angle DFB=\frac 12EF\times FB\times \angle EFB.$$
Notice that we have $\angle EFB=90^\circ$, and $\sin\angle DFB=\sin \angle DAB=DB/AB$. So we only need to prove that
$$\frac{DB}{AB}=\frac{EF}{DF}$$
Notice that $ED$ is tangent, so $\triangle EDF\sim \triangle EAD$, so we have $\frac{EF}{DF}=\frac{DE}{DA}$. Also since $ED$ is tangent, so $\angle EDB=\angle DAB$. Also, we have $\angle ADB=\angle DEB=90^\circ$. So $\triangle EDB\sim \triangle DAB$. Therefore we have $\frac{DE}{DA}=\frac{DB}{AB}$
A: Here is another approach using similar triangles.

As $D$ is midpoint of $ \overset{\huge\frown}{AC}$, $OD$ is perp bisector of $AC$, and as $DE \parallel AC$ and $DH \parallel CE, ~CE = DH$.
So, $HK = \frac 12 CE = \frac 12 DH \implies HK = DK$
Given $\angle DAH = \angle DBE, \triangle DAH \sim \triangle DBE$. As $BJ$ and $AK$ divide $ \overset{\huge\frown}{CD}$ in same ratio, we must have $DJ = JE~$, since we showed above that $~HK = DK$.
