91Ó°ÊÓ

Here A is the midpoint of $\stackrel{¯}{OX}$, $\stackrel{¯}{AB}âˆ\yen \stackrel{¯}{XY}$ and $\stackrel{¯}{BC}âˆ\yen \stackrel{¯}{YZ}$.

In $\mathrm{Δ}XOY$, A is the midpoint of $\stackrel{¯}{OX}$,

And, $\stackrel{¯}{AB}âˆ\yen \stackrel{¯}{XY}$.

So,

B is a midpoint of $\stackrel{¯}{OY}$.

(According to the converse of midpoint theorem, if a line is drawn through the midpoint of a side of a triangle parallel to the other side, then it bisects the third side.)

In $\mathrm{Δ}YOZ$, B is the midpoint of $\stackrel{¯}{OY}$,

Also, $\stackrel{¯}{BC}âˆ\yen \stackrel{¯}{YZ}$.

So,

C is a midpoint of $\stackrel{¯}{OZ}$.

(According to the converse of midpoint theorem, if a line is drawn through the midpoint of a side of a triangle parallel to the other side, then it bisects the third side.)

In $\mathrm{Δ}XOZ$, A is a midpoint of $\stackrel{¯}{OX}$ and C is a midpoint of $\stackrel{¯}{OZ}$.

Also,

According to the midpoint theorem, the line joining the midpoints of two sides of a triangle is parallel to the third side of the triangle.

So,

$\stackrel{¯}{AC}âˆ\yen \stackrel{¯}{XZ}$

Therefore, $\stackrel{¯}{AC}âˆ\yen \stackrel{¯}{XZ}$(proved).