Bisection of a a triangular area

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!

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