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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 $DE$

enter image description here

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 $AC$.

PNT
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2 Answers2

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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 $\frac{DE}{DA}=\frac{DB}{AB}$

JetfiRex
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  • This problem was in “The power of a point” section, so I expected that the solution might have something to do with it. – PNT Feb 21 '22 at 18:33
  • @Yassir That $EDF$ and $EAD$ are silimar can be proved by the power of a point. But I think calculating the angles is enough. – JetfiRex Feb 21 '22 at 19:46
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    Here is an alternative approach, we construct a circle with diameter EB then since $\angle BFE =90$, $F\in \Omega$, wich means . $Pow_{ \Gamma}M=Pow_{\Omega} M$ (M is the intersection of BF with DE). now just by doing some algebra we get $ME=MD$. Is that corret? – PNT Feb 21 '22 at 20:47
  • @Yassir No more algebra is needed since you'll get the equal power, and $MF, MD$ are both tangent, they have to be equal. – JetfiRex Feb 21 '22 at 20:55
  • @Yassir your approach using power of point is nice. As $DE$ is tangent to both $\Gamma$ and $\Omega$ and both circles pass through $F$, we must have $MD = ME$ using power of point $M$ with respect to both circles. – Math Lover Feb 22 '22 at 01:03
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Here is another approach using similar triangles.

enter image description here

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 $~HK = DK$.

Math Lover
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