<s>[INST] Menelaus theorem & collinear points 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. 

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. 
 [/INST] 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 $\angle \, ACK$, 
$$LE = KE$$

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 $AM = MC$. 
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