Euclidean geometry- how do I finish this proof

Question

The problem at hand: In triangle $ABC$, $D$,$E$,$F$ are points on the sides BC,CA,AB. A,B,C are further points on a bigger triangle $XYZ$ such that $EF$ $||$ $YZ$, $FD$ $||$$ZX$, $DE$ $||$ $XY$

Prove that $f(ABC)^2$ = $f(DEF)$$×$ $f(XYZ)$

MY ATTEMPT: There are probably easier ways to do this(correction: there are definitely easier ways to do this) but the way I went about it was,

Let's imagine, D,E,F are the midpoints of the sides of ABC, which are in turn midpoints of the sides of XYZ. The ratio of $f(ABC)$ to $f(DEF)$ is 4, which is the same as the ratio of XYZ to ABC, so the proposition certainly holds for this case. Now, let's say we move any one of the points D,E,F up or down the line on which they are. Let's say, point E is displaced by an angle $x$, the point Y must then be displaced by an angle $x$, to maintain the condition that $XY $||$ $$DE$. Pt. A must then be displaced by an angle $x$, to ensure that it remains on the line $XY$. In this process, all three triangles have gained an extra area. Now, the ratio of the areas ABC and DEF equals $AB^2$:$DE^2$, and the ratios of XYZ to ABC is $XY^2$:$AB^2$. If we can prove that the ratios of the areas of the new triangles that have been created, is the same as the ratio of the original triangles, we have proved that the proposition still holds. Say, $A$ was displaced to $A'$, $Y$ was displaced to $Y'$ and $E$ to $E'$, then the only way to prove that is to prove $AA'B$ similar to $DE'E$ similar to $XY'Y$. Remember we chose point E to be displaced without loss of generality, and we chose an arbitrary angular displacement. So if we can prove this, we can prove that the given proposition holds for all points, because all points on AB,BC,CA can be reached by displacing the midpoints by some amount. Now, proving those 3 triangles to be similar is harder than expected and that's where I need help. Also if I have messed up at any point before in the solution then please tell me too. Thank you

Answer 1

I have concerns about this claim:

Let's say, point E is displaced by an angle $x$, the point Y must then be displaced by an angle $x$, to maintain the condition that $XY $||$ $$DE$.

Issues:
1. When you say "displaced by angle $x$", what is your reference point?
2. When $E$ is displaced, note that (with $A, B, C, D, F$ fixed) points $X, Y, Z$ also are displaced. How do you know how much $XY$ is moved by?
3. Similarly, what is the reference point when you talk about "Y must be displaced by an angle $x$"?

Note: It can be proven that $DX, EY, CZ$ are concurrent at (say) $O$, which might be where you're coming from.
However, note that $O$ depends on the placement of these points. In particular, when $OE$ is displaced by an angle $x$, we get a new $O'$.

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The following claim need not be true:

Now, the ratio of the areas ABC and DEF equals $AB^2$:$DE^2$, and the ratios of XYZ to ABC is $XY^2$:$AB^2$.

This is because $ABC$ and $DEF$ need not be similar (which seems to be the crux of your argument).
$D, E, F$ can be any 3 points on $ABC$, and we can construct a corresponding $XYZ$ (by taking parallel lines through $A, B, C$).