Solution:

We will use a well-known theorem from geometry. A midsegment in a triangle is called a segment that joins the midpoints of two of its sides.

Theorem. The midsegment in a triangle connecting two sides in a triangle is parallel to the third side and half of its length. In other words, if $K$ and $L$ are the midpoints of sides $XZ$ and $YZ$ of $\triangle XYZ$, then the midsegment $KL$ is parallel to side $XY$ and $KL$ is half as long as $XY$.

- As a consequence, the four midpoints $M$, $N$, $P$, and $Q$ in $ABCD$ in our problem form a parallelogram $MNPQ$. Indeed, since $QP$ is parallel to $AC$ (as a midsegment in $\triangle ACD$), and $MN$ is parallel to $AC$ (as a midsegment in $\triangle ACB$), it follows that $QP$ and $MN$ are parallel. They are also half as long as $AC$ and hence equal in length. This means that quadrilateral $MNPQ$ has parallel and equal in length opposite sides, and hence it is a parallelogram.

- From our problem we know that the diagonals $QN$ and $MP$ of this parallelogram $MNPQ$ are equal in length. This means that the parallelogram is actually a rectangle (another famous theorem from geometry). So now we know that $MNPQ$ is a rectangle, i.e., $PN$ and $PQ$ are perpendicular.

- As midsegments in $\triangle ACD$ and $\triangle DBC$, $QP$ and $PN$ are parallel correspondingly to $AC$ and $DB$. This implies that $AC$ and $BD$ are perpendicular to each other, completing our proof.