# Triangles area related problem

The question is :-

In $$\Delta ABC$$ , $$X$$ 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 area of $$\Delta ABC$$ is given by $$\Delta ABC= [(\Delta AXZ)^{1/3} +(\Delta BYZ)^{1/3}]^3$$

I gave this question a good try for at least an hour but I am unable to solve it .

I derived up the following result Let $$\Delta CXZ =a ;$$ $$\Delta CYZ =b ;$$ $$\Delta AXZ =c ;$$ $$\Delta BYZ =d ;$$

Then by applying area theorems and using the given above relation i can easily confer:- $$b^2=a \cdot d ;$$ $$a^2=b \cdot c ;$$

From here we can also get $$a$$ and $$c$$ in terms of $$b$$ and $$d$$ But the main problem is how to relate these to area of the whole triangle since $$\Delta AZB$$ is creating problem(because I am unable to find its area in terms of $$c$$ and $$d$$)

Please tell whether my approach is right and what next should I do to solve the question or if not please the the correct method and approach with proper steps

First, let $$\frac {AX}{XC}=\frac {CY}{YB}=\frac {XZ}{ZY} = k$$ (this is just to make things easier). I also started from $$A(\Delta BYZ) = S$$. Then, except $$\Delta ABZ$$, we could find all other areas without a problem as you suggested. Now, notice that what we need to find is not $$A(\Delta ABZ)$$ but $$A(\Delta ABC)$$. Therefore, draw $$AY$$. Then $$A(\Delta AYZ) = k^2S$$. Then, since $$A(\Delta AYC) = k(k+1)^2S$$, we have $$A(\Delta ABY) = (k+1)^2S$$. Rest is simple algebra.
Let $$S_{\Delta AZX}=a$$, $$S_{\Delta BZY}=b$$ and $$\frac {AX}{XC}=\frac {CY}{YB}=\frac {XZ}{ZY}=k.$$
Thus, $$S_{\Delta CZY}=kb,$$ $$S_{\Delta CZY}=\frac{a}{k},$$ which gives $$\frac{\frac{a}{k}}{bk}=k$$ and $$k=\sqrt[3]{\frac{a}{b}}.$$ But $$\frac{S_{\Delta CXY}}{S_{\Delta ABC}}=\frac{CX\cdot CY}{CA\cdot CB}=\frac{1}{1+\frac{AX}{CX}}\cdot\frac{\frac{CY}{BY}}{1+\frac{CY}{BY}}=\frac{k}{(1+k)^2}.$$ Id est, $$\frac{kb+\frac{a}{k}}{S_{\Delta ABC}}=\frac{k}{(1+k)^2},$$ which gives $$S_{\Delta ABC}=\frac{\left(kb+\frac{a}{k}\right)(1+k)^2}{k}=\frac{b\left(k^2+\frac{a}{b}\right)(1+k)^2}{k^2}=$$ $$=\frac{b(k^2+k^3)(1+k)^2}{k^2}=b(1+k)^3=b\left(1+\sqrt[3]{\frac{a}{b}}\right)^3=\left(\sqrt[3]a+\sqrt[3]b\right)^3$$ and we are done!