TITLE: Prove that $\triangle ABC=\left(\triangle DEF \cdot \triangle XYZ\right)^{1/2}$
QUESTION [5 upvotes]: 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.

REPLY [2 votes]: Say, $\triangle DEF = p$, then triangle $\triangle XYZ = p(t^2)$ where t is the ratio of sides of $\triangle XYZ$ to $\triangle ABC$.
$\triangle XYZ = [XDEY] + [YEFZ] + [XDFZ] + \triangle DEF$ (3 parallelograms + $\triangle DEF$).
Say, $EF = a, FD = b, DE = c$
$\triangle XYZ = \dfrac{1}{2}[c(1+t)h_3 + a(1+t)h_1 + b(1+t)h_2] + \triangle DEF$
$p(t^2) = \dfrac{1}{2}[c(1+t)h_3 + a(1+t)h_1 + b(1+t)h_2] + p$
$2p(t^2) = c(1+t)h_3 + a(1+t)h_1 + b(1+t)h_2 + 2p$ ...(i)
Now, $\triangle ABC = \triangle CDE + \triangle AEF + \triangle BDF + \triangle DEF$
$\triangle ABC = \dfrac{1}{2}(c.h_3 + a.h_2 + b.h_1) + p$ ...(ii)
From (i) and (ii),
$p(t^2) = (\triangle ABC - p)(1+t) + p$
$p(t-1) = \triangle ABC - p$
$\triangle ABC = pt = \sqrt{p.pt^2} = \sqrt{\triangle DEF.\triangle XYZ}$