An altitude is divided into 5 equal parts by 4 lines. Prove that the the areas of alternate sections are equal. The question is as follows :
Let their be a triangle ABC. Make altitude AD on C. Divide this altitude in 5 equal parts with lines EF, GM, IJ, KL intersecting at points M,N,O,P respectively. 
We have to prove that,
area(AEM + GNOI + KPBD + MFHN + SOPL) = area(AFM + HNOS + LPDC + EMNG + IOKP)
This question came in the R.M.O. and was the only question that was causing problems. I unfortunately assumed the lines to be parallel and proved it using similarity. But I have honestly no clue as to how to solve  it now.
Please help.
 A: It would've been better if you had posted an image. Anyway, I'll attempt to answer this question based on what I could make of it.

This is my interpretation of your question. Let the intersecting points of $EF, GH, IJ, KL$ with $AD$ be $M, N, O, P$ respectively. Let the length of each of the 5 segments into which AD is divided be $h$. Also let $EM$ be $x$ and $MF$ be $y$


*

*Now $\triangle AEM \sim \triangle AGN \sim \triangle AIO \sim \triangle AKP \sim \triangle ABD$. Therefore $GN, IO, KP, BD$ are equal to $\ 2x,\ 3x,\ 4x,\ 5x$ respectively.

*Similarly, $\triangle AFM \sim \triangle AHN \sim \triangle AJO \sim \triangle ALP \sim \triangle ACD$. Therefore $ NH, OJ, PL, DC$ are equal to $\ 2y,\ 3y,\ 4y,\ 5y$ respectively.
Now you can directly calculate the pink and purple areas.
$$\begin{align}
\text{Pink Area} &= \frac{xh}{2}+\frac{5xh}{2}+\frac{9xh}{2}+\frac{3yh}{2}+\frac{7yh}{2} \\
&= \frac{(15x+10y)\times h}{2}
\end{align}$$
$$\begin{align}
\text{Purple Area} &= \frac{3xh}{2}+\frac{7xh}{2}+\frac{yh}{2}+\frac{5yh}{2}+\frac{9yh}{2} \\
&= \frac{(10x+15y)\times h}{2}
\end{align}$$
Now, according to the question, $\text{Pink Area} = \text{Purple Area}$
$$\begin{align}
\therefore \frac{(15x+10y)\times h}{2} &= \frac{(10x+15y)\times h}{2} \\
15x+10y&=10x+15y \\
\implies x&=y
\end{align}
$$
So, in order for the area to be equal, $BD$ must be equal to $CD$. $\implies \triangle ABC$ must be isosceles.  
