Three cevians in a triangle create four sub-triangles of area $1$. Find the area of a non-triangular region. 
I'm having trouble proving that all the white and green areas have the same area, from there on we can obtain the answer $1+\sqrt5$ by proving that the inner red triangle points are midpoints.

Let $AD$, $BE$ and $CF$ be our chevians of $\Delta ABC$.
Also, let $AD\cap BE=\{P\}$, $AD\cap CF=\{R\}$ and $BE\cap CF=\{Q\}.$
Thus, $$S_{\Delta APE}=S_{\Delta BQF}=S_{\Delta CRD}=S_{\Delta PQR}=1.$$
Let $S_{AFQP}=x,$ $S_{BDRQ}=y$ and $S_{CEPR}=z$.
Thus, $$S_{\Delta BDR}=\frac{S_{\Delta BDR}}{S_{\Delta CDR}}=\frac{BD}{DC}=\frac{x+y+2}{z+2},$$
which gives
$$S_{\Delta BQR}=y-\frac{x+y+2}{z+2}=\frac{yz+y-x-2}{z+2}.$$
Now, $$S_{\Delta AFQ}=\frac{S_{\Delta AFQ}}{S_{\Delta BFQ}}=\frac{AF}{BF}=\frac{x+z+2}{y+2},$$
which gives
$$S_{\Delta AQP}=x-\frac{x+z+2}{y+2}=\frac{xy+x-z-2}{y+2}.$$
Thus, $$\frac{\frac{x+z+2}{y+2}+1}{\frac{xy+x+z-2}{y+2}}=\frac{S_{\Delta AFQ}+S_{\Delta BQF}}{S_{\Delta AQP}}=\frac{S_{\Delta ABQ}}{S_{\Delta AQP}}=$$
$$=\frac{BQ}{PQ}=\frac{S_{\Delta BQR}}{S_{\Delta PQR}}=S_{\Delta BQR}=\frac{yz+y-x-2}{z+2}.$$
Id est, $$\frac{x+y+z+4}{xy+x-z-2}=\frac{yz+y-x-2}{z+2}$$ or
$$(y+1)(z^2+x^2-xy+2x+4z-xyz+4)=0,$$ which gives
$$z^2+x^2-xy+2x+4z=xyz-4.$$
By the same way we can obtain:
$$x^2+y^2-yz+2y+4x=xyz-4$$ and
$$y^2+z^2-zx+2z+4y=xyz-4.$$
Now, let $x=\max\{x,y,z\}$.
Thus, $$x^2+y^2-yz+2y+4x-(y^2+z^2-zx+2z+4y)=0$$ or
$$(x-z)(x+z)+z(x-y)+4x-2y-2z=0$$ or
$$(x-z)(x+z+2)+(x-y)(z+2)=0,$$ which gives $$x=y=z,$$
$$x^2+6x=x^3-4$$ or
$$x^3-x^2-6x-4=0$$ or
$$x^3+x^2-2x^2-2x-4x-4=0$$ or
$$(x+1)(x^2-2x-4)=0$$ or
$$x=1+\sqrt5$$ and we are done!
