Solution:

Let $[\mathcal{P}]$ denote the area of a polygon $\mathcal{P}$.

The important claim is that if $\overline{K L} \parallel \overline{M N}$, then $[K L M] = [K L N]$. This is a simple consequence of the formula $A = \frac{1}{2} b h$.

Then, we find that
$$\begin{aligned} & [A B C] = [X B C] \quad (\text{since } \overline{A X} \parallel \overline{B C}) \\ & = [X Y C] \quad (\text{since } \overline{B Y} \parallel \overline{X C}) \\ & = [X Y Z] \quad (\text{since } \overline{C Z} \parallel \overline{X Y}) \end{aligned}$$
as desired.