Answer
$M_{AC}=M_{BD}$, so the point $\displaystyle \left(\frac{s}{2},\frac{s}{2}\right)$ is the midpoint of both diagonals.
In other words, the diagonals of a square intersect at their midpoints.
Work Step by Step
Using the hint, let
$A=(0,0),B=(0,s), C=(s,s),$ and $D=(s,0)$ be the vertices of the square.
The diagonals are $\overline{AC}$ and $\overline{BD}$ . Midpoints:
$M_{AC}=\displaystyle \left(\frac{0+s}{2},\frac{0+s}{2}\right)=\left(\frac{s}{2},\frac{s}{2}\right)$
$M_{BD}=\displaystyle \left(\frac{0+s}{2},\frac{s+0}{2}\right)=\left(\frac{s}{2},\frac{s}{2}\right)$
$M_{AC}=M_{BD}$, so the point $\displaystyle \left(\frac{s}{2},\frac{s}{2}\right)$ is the midpoint of both diagonals.
In other words, the diagonals of a square intersect at their midpoints.
