Triangles area related problem

Question

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

Answer 1

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.

Answer 2

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!