Bisection of a a triangular area Here's the sketch:

From the inner point P of a triangle ABC the three connecting lines to the corner points are drawn. In addition, the lines PE, PD and PF are each drawn parallel to a median of ABC. Show that the grey areas cover half of the triangular area.
I thought about using ceva's theorem, but I don't really know how to start... any ideas?
 A: Ronel Leker's solution.
Let $\{F_1,F_2\}\subset AC$, $\{E_1,E_2\}\subset BC$ and $\{D_1,D_2\}\subset AB$ such that
$F_1E_2||AB$, $D_2E_1||AC$, $D_1F_2||BC$ and $F_1E_2\cap D_2E_1\cap D_1F_2=\{P\}.$
Thus, since $$\Delta PE_2E_1\sim\Delta F_1PF_2\sim\Delta D_2D_1P\sim\Delta ABC,$$ we obtain that $PE$, $PF$ and $PD$ are medians of $\Delta PE_1E_2,$ $\Delta PF_1F_2$ and $\Delta PD_1D_2$ respectively.
Also, since $AF_1PD_2$, $BD_1PE_2$ and $CE_1PF_2$ are parallelograms, we obtain:
$$S_{\Delta PAD}+S_{\Delta PBE}+S_{\Delta PCF}=$$
$$=\left(S_{\Delta PAF_1}+S_{\Delta PDD_1}\right)+\left(S_{\Delta PBD_1}+S_{\Delta PEE_1}\right)+\left(S_{\Delta PCE_1}+S_{\Delta PFF_1}\right)=$$
$$=\left(S_{\Delta PAF_1}+S_{\Delta PFF_1}\right)+\left(S_{\Delta PBD_1}+S_{\Delta PDD_1}\right)+\left(S_{\Delta PCE_1}+S_{\Delta PEE_1}\right)=$$
$$=S_{\Delta PAF}+S_{\Delta PBD}+S_{\Delta PCE}$$ and we are done!


A: COMMENT:
Let the perpendiculars from p on AB , BC and AC be $i, j , k$ respectively. Using relation:
$\frac{i}{h_C}+\frac{j}{h_B}+\frac{k}{h_A}=1$
where $h_c$, $h_B$ and $h_A$ are hights of triangle from vertices C, B and A respectively; we can find following relations:
$S_{ABC}=S_{PAD}.\frac{AB}{AD}+S_{PFC}.\frac{AC}{FC}+S_{PEB}.\frac{BC}{EB}$
$S_{ABC}=S_{PDB}.\frac{AB}{DB}+S_{PFA}.\frac{AC}{AF}+S_{PEc}.\frac{BC}{EC}$
If P is the intersection of medians then the question is answered. If P is not intersection of medians then we conclude that RHS of these relations is symmetric and therefore they must be equal, hence the question is answered.
A: The way I would approach this problem is to do an affine transformation into an equilateral triangle, then use Carnot's theorem$$AD^2 - DB^2 + BE^2 - EC^2 + CF^2 - FA^2 = 0$$to get$$AD - DB + BE - EC + CF - FA = 0,$$which is the necessary and sufficient condition for the existence of $P$. Then we just have to express the ratio $PD:PE:PF$ in terms of the above lengths (which frankly I did not bother, but know it is possible). It is probably equivalent to darij grinberg's solution in the comments.
Because this and darij grinberg's solution makes use of the non-affine property of equilateral triangles, while the question is affine-invariant, the solution seems "morally wrong". That is why Ronel Leker's solution, by way of Michael Rozenberg here, seems more elegant. I also think that in darij grinberg's solution, observing that the expression is quadratic without actually computing the messy algebra is very elegant.
