A quadrilateral ABCD's diagonals are perpendicular. What is the name of the quadrilateral formed by joining the midpoints of its sides?

Given, AC, and BD are diagonals of a quadrilateral ABCD are perpendicular.

Points P, Q, R and S are mid-points of AB, BC, CD and DA respectively.

In$∆\mathrm{ABC}$, P and Q are mid-points of AB and BC respectively.

So, by the Mid-point theorem,

$$\begin{gathered} \therefore \mathrm{PQ}\parallel \mathrm{AC} ; \\ \mathrm{PQ}=\frac{1}{2}\mathrm{AC} ............\left(1\right)(\mathrm{Mid} \mathrm{point} \mathrm{theorem}) \end{gathered}$$

Similarly, in $∆\mathrm{ACD}$

R and S are the mid-points of sides CD and AD respectively.

$$\begin{gathered} \therefore \mathrm{SR}\parallel \mathrm{AC}; \\ \mathrm{SR}=\frac{1}{2}\mathrm{AC} .......\left(2\right)(\mathrm{Mid} \mathrm{point} \mathrm{theorem}) \end{gathered}$$

From (1) and (2), we get

$\mathrm{PQ}\parallel \mathrm{SR} \mathrm{and} \mathrm{PQ}=\mathrm{SR}$

Hence, PQRS is a parallelogram (pair of opposite sides are equal and parallel)

Now, $\mathrm{RS}\parallel \mathrm{AC} \mathrm{and} \mathrm{QR}\parallel \mathrm{BD}$

Also, $\mathrm{AC}\perp \mathrm{BD}$ (Given)

$\therefore \mathrm{RS}\perp \mathrm{QR}$

Thus, PQRS is a rectangle.