<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] 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.
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