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.
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.