Proof: Quadrilateral formed by joining midpoints of a rhombus is a rectangle.

Given: A rhombus ABCD with midpoints E, F, G, and H of its sides AB, BC, CD, and DA, respectively.

To prove: Quadrilateral EFGH is a rectangle.

Proof:

Step 1: Show that the opposite sides of EFGH are parallel.

Since E is the midpoint of AB and F is the midpoint of BC, we can conclude that EF is parallel to the side AD of the rhombus because opposite sides of a rhombus are parallel.

Similarly, since F is the midpoint of BC and G is the midpoint of CD, we can conclude that FG is parallel to AB.

By the same reasoning, GH is parallel to BC, and EH is parallel to CD.

Thus, we have established that the opposite sides of EFGH are parallel.

Step 2: Show that the opposite sides of EFGH are equal in length.

Since E is the midpoint of AB and H is the midpoint of AD, we can conclude that EH is equal to half the length of AD.

Similarly, FG is equal to half the length of BC, EF is equal to half the length of AB, and GH is equal to half the length of CD.

Since a rhombus has all sides equal in length, we can conclude that EH = FG = EF = GH.

Thus, we have established that the opposite sides of EFGH are equal in length.

Step 3: Show that the diagonals of EFGH bisect each other.

The diagonals of EFGH are EG and FH. Since EG is the line segment joining the midpoints of the diagonals of the rhombus ABCD, we can conclude that EG is equal to half the length of AC.

Similarly, FH is equal to half the length of AC.

Since EG = FH, we can conclude that the diagonals of EFGH bisect each other.

Step 4: Conclude that EFGH is a rectangle.

Since EFGH has opposite sides parallel, opposite sides equal in length, and diagonals that bisect each other, we can conclude that EFGH is a rectangle.

Hence, the quadrilateral formed by joining the midpoints of a rhombus is a rectangle.

This completes the proof.