$X$ and $Y$ are points on the side $LN$ of the triangle $LMN$ such that $LX = XY = YN$. Through $X$,a line is drawn parallel to $LM$ to meet $MN$ at $Z$ (See figure). Prove that $\operatorname{ar}(LZY) = \operatorname{ar}(MZYX)$.

(N/A) We have to prove that $\operatorname{ar}(\triangle LZY) = \operatorname{ar}(MZYX)$.
Since $\triangle LXZ$ and $\triangle MXZ$ are on the same base $XZ$ and between the same parallels $LM$ and $XZ$,we have:
$\operatorname{ar}(\triangle LXZ) = \operatorname{ar}(\triangle MXZ) \quad \dots(1)$
Adding $\operatorname{ar}(\triangle XYZ)$ to both sides of $(1)$,we get:
$\operatorname{ar}(\triangle LXZ) + \operatorname{ar}(\triangle XYZ) = \operatorname{ar}(\triangle MXZ) + \operatorname{ar}(\triangle XYZ)$
$\Rightarrow \operatorname{ar}(\triangle LZY) = \operatorname{ar}(MZYX)$.