<s>[INST] Area of a triangle in terms of areas of certain subtriangles In triangle $ABC$ , $X$ and $Y$ are points on sides $AC$ and $BC$ respectively . If $Z$ is on the segment $XY$ such that $\frac{AX}{XC} = \frac{CY}{YB} = \frac{XZ}{ZY}$ , then how to prove that the area of  triangle $ABC$ is given by $[ABC]=([AXZ]  ^{1/3}  + [BYZ]^{1/3})^3.$
 [/INST] We can say that points $X$, $Y$, $Z$ separate respective segments $AC$, $CB$, $XY$ into sub-segments with ratios $b:r$ (think "blue : red"), where $b+r=1$.

Let's add a few perpendiculars, transfer our ratio-colors accordingly (as the reader may verify, using similar triangles), and compute the area of $\triangle AXZ$:

$$\begin{align}
|\triangle AXZ| &= \frac{1}{2}\;|\overline{AX}|\;|\overline{RZ}| \\[6pt]
&= \frac{1}{2}\cdot b\;|\overline{AC}|\cdot b\;|\overline{RS}| \\[6pt]
&= \frac{1}{2}\cdot b\;|\overline{AC}|\cdot b\cdot b\;|\overline{PB}| \\[6pt]
&= b^3\cdot \frac{1}{2}|\overline{AC}|\;|\overline{PB}| \\[6pt]
&= b^3\;|\triangle ABC| \\[8pt]
\text{Similarly, } |\triangle BYZ| &= r^3\;|\triangle ABC|
\end{align}$$
Therefore,
$$\frac{|\triangle AXZ|}{|\triangle ABC|} = b^3 \qquad\text{and}\qquad 
\frac{|\triangle BYZ|}{|\triangle ABC|} = r^3$$
Recalling $b+r=1$, the result follows. $\square$
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