Hello, Joanne!

Prove that the figure formed by connecting the adjacent midpoints of any rectangle is a rhombus.

First, make a sketch . . . Code:

P
A * - - - - - * - - - - - * B
| * * |
| * * |
| * * |
S * * Q
| * * |
| * * |
| * * |
D * - - - - - * - - - - - * C
R

Let $\displaystyle ABCD$ be the rectangle and $\displaystyle P,Q,R,S$ be the midpoints.

There are right triangles in each corner.

We can prove that they are all congruent.

Therefore: .$\displaystyle PQ = QR = RS = SP$ . and $\displaystyle PQRS$ is a rhombus.

For example, consider $\displaystyle \Delta PAS$ and $\displaystyle \Delta PBQ$

$\displaystyle AD = BC$ . . . Opposite sides of a rectangle are equal.

$\displaystyle S$ is the midpoint of $\displaystyle AD$; $\displaystyle Q$ is the midpont of $\displaystyle BC.$

. . $\displaystyle AS = \tfrac{1}{2}AD,\;BQ = \tfrac{1}{2}BC \quad\Rightarrow\quad AS = BQ$

$\displaystyle \angle A \:=\:\angle B \:=\:90^o$

$\displaystyle P$ is the midpont of $\displaystyle AB \quad\Rightarrow\quad AP = PB$

Hence: .$\displaystyle \Delta PAS \cong \Delta PBQ\;\;\text{(s.a.s.)}$

Therefore: .$\displaystyle PS = PQ$

Continue in the same manner and prove all the hypotenuses are equal.