What can we say about the areas of these two triangles?

Question

Given triangle ABC. Let X be a point on AB, Y be a point on BC and Z be a point on AC. Now suppose we reflect X, Y, Z around the midpoint of the sides they are on and label the images X', Y' and Z'. Is it always true that triangle XYZ and triangle X'Y'Z' have the same areas?

Answer 1

Renaming your points (I like $X$ opposite $A$, etc), I'll define three ratios $$x := \frac{|BX|}{|BC|} \qquad y := \frac{|CY|}{|CA|} \qquad z := \frac{|AZ|}{|AB|}$$

Then note that, for instance,

$$|\triangle AYZ| = \frac12\;|AY||AZ|\sin A = \frac12 \;(1-y) |CA|\; z|AB|\;\sin A=z(1-y)|\triangle ABC|$$ and likewise for the other sub-triangles at the vertices. Therefore, $$\begin{align} |\triangle XYZ| &= |\triangle ABC| - |\triangle AYZ| - |\triangle XBZ| - |\triangle XYC| \\ &= |\triangle ABC|\;\left(\; 1 - (1-y) z - ( 1 - z ) x - ( 1 - x ) y\;\right) \\[4pt] &= |\triangle ABC|\;\left(\; 1 - x - y - z + y z + z x + x y \;\right) \\[4pt] &= |\triangle ABC|\;\left(\; (1-x)(1-y)(1-z) + x y z \;\right) \\[4pt] \end{align}$$

This result is preserved if we swap ratios $$x \leftrightarrow (1-x) \qquad y \leftrightarrow (1-y) \qquad z \leftrightarrow (1-z)$$ which is precisely what happens in the computation of the area of $\triangle X^\prime Y^\prime Z^\prime$. $\square$