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From vertex C of the right triangle ABC height CK is dropped and in triangle ACK bisector CE is drawn. Line that passes through point B parallel to CE meets CK at point F. Prove that line EF divides segment AC in halves.

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So far I have:

Construct point M on AC such that AM=MC. WE want to show that M, E, and F are collinear. So we want to prove using Menelaus' that $\displaystyle\frac{AM}{CM} \cdot \displaystyle\frac{CF}{KF} \cdot \displaystyle\frac{KE}{AE} = 1$.

We know AM=MC so that cancels out to 1. From the angle bisector theorem in triangle ACK we know that $\displaystyle\frac{KE}{AE} = \displaystyle\frac{CK}{CA}$.

So from this we have $\displaystyle\frac{CF}{KF} \cdot \displaystyle\frac{CK}{CA}=1$. Using the fact that $\triangle CKE$ is similar to $\triangle BFK$ we know $\displaystyle\frac{CK}{KF}=\displaystyle\frac{EK}{KB}$

Now we obtain that $\displaystyle\frac{CF}{CA} \cdot \displaystyle\frac{EK}{KB} = 1$

FROM HERE I AM STUCK.

Jack D'Aurizio
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  • yes it was preferred we use Menelaus' theorem as this is what we are learning –  Mar 14 '17 at 22:22
  • All right, much better now (I took the liberty to add a diagram). – Jack D'Aurizio Mar 14 '17 at 22:30
  • great...that is somewhat similar to the diagram that i drew...do you know where i can go from what i have done so far? or have i made an error somewhere? –  Mar 14 '17 at 22:34

2 Answers2

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You have to check that by choosing $M$ as the midpoint of $AC$ then $$ \frac{AM}{CM}\cdot\frac{CF}{KF}\cdot\frac{KE}{AE} = 1 $$ holds, but $\frac{AM}{CM}=1$ and by the bisector theorem $\frac{KE}{AE}=\frac{CK}{CA}$, so it is enough to prove that $$ CF\cdot CK = AC\cdot KF $$ or $$ \frac{CA}{CK} = \frac{FC}{FK} = 1+\frac{CK}{FK}=1+\frac{EK}{KB}. $$ If $a,b$ are the lengths of the legs of $ABC$, we have $CK=\frac{ab}{\sqrt{a^2+b^2}},CA=b$, $AK=\frac{b^2}{\sqrt{a^2+b^2}}$, $BK=\frac{a^2}{\sqrt{a^2+b^2}}$ and $$AE=\frac{b}{b+\frac{ab}{\sqrt{a^2+b^2}}}\cdot\frac{b^2}{\sqrt{a^2+b^2}}=\frac{b^2}{a+\sqrt{a^2+b^2}}$$ $$EK=\frac{\frac{ab}{\sqrt{a^2+b^2}}}{b+\frac{ab}{\sqrt{a^2+b^2}}}\cdot\frac{b^2}{\sqrt{a^2+b^2}}=\frac{ab^2}{a^2+b^2+a\sqrt{a^2+b^2}}$$ The problem boils down to tedious&straightforward algebra.

Jack D'Aurizio
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Let the line through point $E$ and orthogonal to $CA$ intersect $CA$ at point $L$. Let the line through point $F$ and orthogonal to $BC$ intersect $BC$ at point $N$. Since $AB$ is orthogonal to $CK$ and $CE$ is the angle bisector of angle $\angle \, ACK$, $$LE = KE$$

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Let $\angle \, BAC = \alpha = \angle \, CAK$. Then $\angle \, ACK = 90^{\circ} - \alpha$ and $\angle \, BCK = \alpha$ which means that $$\angle \, ACE = \frac{1}{2} \, \angle \, ACK = 45^{\circ} - \frac{\alpha}{2}$$ since $CE$ is the angle bisector of angle $\angle \, ACK$. Consequently $$\angle \, BEC = \angle \, ACE + \angle \, CAE = 45^{\circ} - \frac{\alpha}{2} + \alpha = 45^{\circ} + \frac{\alpha}{2}$$ Since $BF$ is parallel to $CE$ $$\angle \, EBF = \angle \, BEC = 45^{\circ} + \frac{\alpha}{2}$$ However, $$\angle \, EBN = 180^{\circ} - \angle \, CBA = 180^{\circ} - (90^{\circ} - \angle \, BAC) = 90^{\circ} + \alpha$$ and so $$\angle \, NBF = \angle \, EBN - \angle \, EBF = 90^{\circ} + \alpha - 45^{\circ} - \frac{\alpha}{2} = 45^{\circ} + \frac{\alpha}{2} = \angle \, EBF$$ which means that $BF$ is the angle bisector of $\angle \, EBN$ and thus $$NF = KF$$ Since triangles $AEL$ and $CFN$ are right and $$\angle \, EAL = \angle \, BAC = \alpha = \angle \, BCK = \angle \, FCN$$ the two triangles $AEL$ and $CFN$ are similar and thus $$\frac{LE}{EA} = \frac{NF}{FC}$$ Take point $P$ on line $CF$ so that $F$ is the midpoint of $CP$. Then $FC = FP$. Therefore, $$\frac{KE}{EA} = \frac{LE}{EA} = \frac{NF}{FC} = \frac{KF}{FP}$$ which by the Thales' intercept theorem implies that $EF$ is parallel to $AP$. Since $M$ is the intersection point of $EF$ with $CA$, the segment $FM$ is parallel to $AP$ and $F$ is the midpoint of $CP$ by construction, so $M$ is the midpoint of $AC$.

Edit. To use Menelaus' theorem, simply get to the point where we have proven that $$\frac{LE}{EA} = \frac{NF}{CF}$$ as explained above. After that use the fact that $LE = KE$ and $NF = FK$ and apply Menelaus' theorem as follows
$$1 = \frac{CF}{FK} \cdot \frac{KE}{EA} \cdot \frac{AM}{MC} = \frac{CF}{NF} \cdot \frac{LE}{EA} \cdot \frac{AM}{MC}$$ However $$\frac{LE}{EA} = \frac{NF}{CF}$$ is equivalent to $$ \frac{CF}{NF} \cdot \frac{LE}{EA} = 1$$ which reduces the identity $$1 = \frac{CF}{NF} \cdot \frac{LE}{EA} \cdot \frac{AM}{MC}$$ to $$1 = \frac{AM}{MC}$$ which is possible if and only if $AM = MC$.

Futurologist
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