In $\Delta ABC$ and $\Delta XYZ$,$m \angle A = m \angle X$ and $m \angle B = m \angle Y$. $\overline{AM}$ is a median of $\Delta ABC$ and $\overline{XP}$ is a median of $\Delta XYZ$. Prove that $AM \times YZ = XP \times BC$.

(N/A) $1$. In $\Delta ABC$ and $\Delta XYZ$,we are given $m \angle A = m \angle X$ and $m \angle B = m \angle Y$. By the $AA$ (Angle-Angle) similarity criterion,$\Delta ABC \sim \Delta XYZ$.
$2$. Since the triangles are similar,the ratio of their corresponding sides is equal: $\frac{AB}{XY} = \frac{BC}{YZ} = \frac{AC}{XZ}$.
$3$. From $\frac{BC}{YZ} = \frac{AB}{XY}$,we can write $\frac{BC}{AB} = \frac{YZ}{XY}$.
$4$. Consider $\Delta ABM$ and $\Delta XYP$. We have $m \angle B = m \angle Y$ and $\frac{AB}{XY} = \frac{BM}{YP}$ (since $BM = \frac{1}{2} BC$ and $YP = \frac{1}{2} YZ$,the ratio of halves is equal to the ratio of the sides).
$5$. By the $SAS$ (Side-Angle-Side) similarity criterion,$\Delta ABM \sim \Delta XYP$.
$6$. Therefore,the ratio of their corresponding sides is equal: $\frac{AM}{XP} = \frac{AB}{XY} = \frac{BM}{YP}$.
$7$. Since $\frac{AM}{XP} = \frac{BC}{YZ}$ (because $\frac{AB}{XY} = \frac{BC}{YZ}$),we have $AM \times YZ = XP \times BC$. Hence proved.