In $\triangle ABC$, $X$ and $Y$ are points on the sides $AC$ and $BC$ respectively. If $Z$ is on the segment $XY$ such that $\frac{AX}{XC}=\frac{CY}{YB}=\frac{XZ}{ZY}$, prove that the area of $\triangle ABC$ is given by: $$ \text{area of }\triangle ABC=\left((\text{area of }\triangle AXZ)^{1/3} +(\text{area of }\triangle BYZ)^{1/3}\right)^3 $$
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Your question is exactly the same as problem #836875. – Mick Jun 18 '14 at 09:14