In $\Delta XYZ$,points $A, B, C, D, E, F,$ and $G$ lie on the side $YZ$ such that $YA = AB = BC = CD = DE = EF = FG = GZ$. Prove that $ar(XBE) = \frac{3}{8} ar(XYZ)$.

(N/A) $1$. Let the base $YZ$ of $\Delta XYZ$ be divided into $8$ equal parts by the points $A, B, C, D, E, F,$ and $G$.
$2$. Since $YA = AB = BC = CD = DE = EF = FG = GZ$,we can say that each segment is equal to $\frac{1}{8}$ of the total length $YZ$.
$3$. The area of a triangle is given by the formula $ar = \frac{1}{2} \times \text{base} \times \text{height}$.
$4$. All triangles $\Delta XYZ, \Delta XBE,$ etc.,share the same vertex $X$ and lie on the same base line $YZ$,meaning they all have the same altitude $h$.
$5$. The base of $\Delta XBE$ is $BE$. Since $BE = BC + CD + DE$,and each segment is $\frac{1}{8} YZ$,we have $BE = \frac{1}{8} YZ + \frac{1}{8} YZ + \frac{1}{8} YZ = \frac{3}{8} YZ$.
$6$. Therefore,$ar(XBE) = \frac{1}{2} \times BE \times h = \frac{1}{2} \times (\frac{3}{8} YZ) \times h$.
$7$. This simplifies to $ar(XBE) = \frac{3}{8} \times (\frac{1}{2} \times YZ \times h) = \frac{3}{8} ar(XYZ)$.