In $\triangle ABC$ , $D$, $E$, $F$ are points on the sides $BC$, $CA$, $AB$. Also, $A$, $B$, $C$ are points on $YZ$, $ZX$, $XY$ of $\triangle XYZ$ for which $EF \parallel YZ$, $FD \parallel ZX$, $DE \parallel XY$. Prove that area of $$\triangle ABC=\left(\triangle DEF \cdot \triangle XYZ\right)^{1/2}$$
I really have no idea how to approach this question. Any help would be greatly appreciated. The only thing that I know is $\triangle DEF \sim \triangle XYZ$.
I do not know about homothety and am expected to only solve this problem using elementary techniques such as similarity, Menelaus theorem, Ceva Theorem etc. Trigonometry is also allowed.
Best Answer
Without loss of generality, suppose that $[XYZ]$ (the area of $\triangle XYZ$) is $1$, and the ratio of similarity between $\triangle DEF$ and $\triangle XYZ$ is $r<1$ (so that $[DEF] = r^2$).
Let $a, b, c$ be the distances between $EF$ and $YZ$, between $ZX$ and $FD$, and between $XY$ and $DE$, respectively.
Then we have $[AEF] = \frac a2 \cdot EF$, $[BFD] = \frac b2 \cdot FD$, and $[CDE] = \frac c2 \cdot DE$ by the formula for triangle area; adding them together, we have $$[ABC] - [DEF] = \frac a2 \cdot EF + \frac b2 \cdot FD + \frac c2 \cdot DE.$$
On the other hand, we have $[AEY] = \frac a2 \cdot AY$, $[AFZ] = \frac a2 \cdot AZ$, $[BFZ] = \frac b2 \cdot BZ$, $[BDX] = \frac b2 \cdot BX$, $[CDX] = \frac c2 \cdot CX$, and $[CEY] = \frac c2 \cdot CY$; adding them together and noting that for example $YZ = AY + AZ$, we have $$[XYZ] - [ABC] = \frac a2 \cdot YZ + \frac b2 \cdot ZX + \frac c2 \cdot XY.$$
Because $r$ is the ratio of similarity between $\triangle DEF$ and $\triangle XYZ$, we have $EF = r \cdot YZ$, $FD = r \cdot ZX$, and $DE = r \cdot XY$, which tells us that
$$ [ABC] - [DEF] = r([XYZ] - [ABC]). $$ Recall that we assumed $[XYZ] = 1$ and $[DEF] = r^2$, so we now have $[ABC] - r^2 = r(1 - [ABC])$. Solving, we get $[ABC] = r$, so $[ABC] = \sqrt{r^2 \cdot 1} = \sqrt{[DEF] \cdot [XYZ]}$.