Triangles area question This question came in RMO, an olympiad in India. I solved it but with the assumption that the lines are parallel, though we are not given this info in the question.

In acute $\triangle ABC$, let D be the foot of perpendicular from A on BC. Consider points K, L, M on segment AD such that AK= KL= LM= MD. Suppose the sum of the areas of the shaded region equals the sum of the areas of the unshaded regions in the following picture. Prove that BD= DC. 

Taken from India Regional Mathematical Olympiad 2014 Question 1 of Region 1

Please help, thank you. This is not a homework question, just saying.
 A: Assuming the lines are parallel with the base of the triangle.
Let's give names to the horizontal line segments in this fashion (forgive the poor visualization):
      A
   a     e
   b     f
   c     g
B  d  D  h  C

Assuming the lines are parallel, and if we call the length of $AK = x$,
the size of the shaded area is:
$$ \frac{ax}{2} + \frac{(b+c)x}{2} + \frac{(e+f)x}{2} + \frac{(g+h)x}{2} = x\frac{a+b+c+e+f+g+h}{2} $$
and the size of the unshaded area is:
$$ \frac{ex}{2} + \frac{(a+b)x}{2} + \frac{(c+d)x}{2} + \frac{(f+g)x}{2} = x\frac{a+b+c+d+e+f+g}{2} $$
Since the description says these areas are equal,
we have:
$$ x\frac{a+b+c+e+f+g+h}{2} = x\frac{a+b+c+d+e+f+g}{2} $$
$$ a+b+c+e+f+g+h = a+b+c+d+e+f+g $$
$$ h = d $$
If we cannot make the assumption of parallel lines,
then I need to think harder.
Probably that's the whole point:
proving that if the shaded and unshaded areas are equal,
then the lines are inevitably parallel with the base,
and therefore $BC = BD$.
A: There is no way to prove that $\overline {BD}$ = $\overline {DC}$ without the assumption that the lines at K, L, and M are parallel to $\overline {BC}$
There simply is not enough information if we say that lines K, L, and M are not parallel.
to Support this answer I point to the Forum on the site where the question resides.
Dividing the altitude of a triangle into four equal parts
The Forum Post is labeled assuming that the lines are parallel even.
I would say that the assumption that the lines are parallel is a valid assumption given all the information that is available.
some posts from the forum:

@Bunny da :P : Are we given that those small triangles are similar? Or any such extra information about those small triangles?
Umm, yeah, you're given that those lines are parallel (though that wasn't explicitly stated in the paper). :P

A: The OP has stressed that the problem doesn't explicitly assume that the lines that pass through $AD$ at $K$, $L$, and $M$ are parallel, so that's the only issue I'm addressing in this answer.
I don't think the result is true if you don't make that assumption.  In particular, it's possible to create an extreme case where $B=D$:  Let the line at $L$ be perpendicular to $AD$, but slant the lines at $K$ and $M$ so that each shaded region has area equal to the unshaded region just below it.  (The line from $K$ will slant down to intersect the perpendicular from $L$ exactly on $AC$, producing two triangles of equal area.)
Added later:  If considering the extreme case isn't convincing, consider what happens if you shave a little bit from the left side of the (isosceles) triangle by moving to a point $B'$ just to the right of $B$.  It's easy to see that this shaves off more shaded region than unshaded, creating a momentary imbalance with more unshaded area than shaded in what remains.  But now if you tilt any of the three lines through $K$, $L$, or $M$ to an appropriate (slight) angle, you can clearly restore the balance.  So unless you make some assumption about those lines (e.g., that they are all parallel, or all perpendicular to $AD$), you cannot draw the conclusion that the problem requires.
