What is this problem even asking for? And what is the answer?

I think it asks for the ratio $a=KH/HM= MN/NL$

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Or is it $\frac{KH}{HM}=\frac{LN}{MN}$?

franzliszt wrote: Or is it $\frac{KH}{HM}=\frac{LN}{MN}$? I doubt for that, because in that case, we would have $HN\parallel KL \parallel BC$ and $HN$ couldn't pass through $B$

Okay. Because if it were the case that $\frac{KH}{HM}=\frac{LN}{MN}$, then I believe the only answer is if $H=N=M$ which would be strange.

This problem is so ambiguous. I will reword it so that there are not four different answers. Quote: In $\triangle ABC$, let $D,E,F$ be the midpoints of sides $BC,CA,AB$, respectively. A line through vertex $A$ cuts segments $FE$ and $DE$ at points $P$ and $Q$ respectively. Given that $\frac{FP}{PE}=\frac{EQ}{QD}$, compute this ratio.

Now here is the solution. We use barycentric coordinates with reference triangle $ABC$. Set $A=(1,0,0),B=(0,1,0),C=(0,0,1)$ and we can easily find $D=(0,1/2,1/2),E=(1/2,0,1/2),F=(0,1/2,1/2)$. Points are segment $FE$ are paramiterized by $(1/2,t,1/2-t)$ from some $t\in\mathbb{R}$. Points on segment $DE$ are paramiterized by $s,1/2-s,1/2)$ for some $s\in\mathbb{R}$. But since $\frac{FP}{PE}=\frac{EQ}{QD}$, we must have $s=t$. Since $A,P,Q$ are colinear, we must have $$\begin{vmatrix}1&0&0\\ 1/2&t&1/2-t\\ t&1/2-t&1/2\end{vmatrix}=0 \Rightarrow \frac{t}2-(1/2-t)^2=0 \Rightarrow 4t^2-6t+1=0 \Rightarrow t=\frac{3\pm\sqrt5}4.$$Since $P$ and $Q$ are on their respective segments (and not the extensions), we take the negative root. Now all that remains is to compute $$\frac{FP}{PE}=\frac{\frac12-t}{t}=\frac{\frac12-\left(\frac{3\pm\sqrt5}4\right)}{\frac{3\pm\sqrt5}4}=\frac{-1+\sqrt5}{3-\sqrt5}=\frac{1+\sqrt5}{2}$$which is the Golden Ratio.