91Ó°ÊÓ

The given diagram is:

The theorem 5-3 states that diagonals of a parallelogram bisect each other.

It is being given that quadrilateral ABCD is a parallelogram.

From the given diagram it can be noticed that $\stackrel{¯}{AC}$ and$\stackrel{¯}{BD}$ are the diagonals of the parallelogram ABCD.

Therefore, by using the theorem 5-3 it can be said that the diagonals $\stackrel{¯}{AC}$and $\stackrel{¯}{BD}$will bisect each other.

From the given diagram it can be noticed that the diagonals $\stackrel{¯}{AC}$ and $\stackrel{¯}{BD}$ are bisecting each other at X.

Therefore, X is the midpoint of $\stackrel{¯}{AC}$.

Therefore, by using the definition of midpoint it can be said that:

$AX=XC$

From the given diagram it can be noticed that:

$AX+XC=AC$

Therefore, it can be obtained that:

$\begin{array}{c}AX+XC=AC\\ AX+AX=AC\\ 2AX=AC\\ AX=\frac{1}{2}AC\end{array}$

Therefore,$AX=\frac{1}{2}AC$

Therefore, the theorem that justifies the statement $AX=\frac{1}{2}AC$is theorem 5-3 which states that diagonals of a parallelogram bisect each other.