The secret to answering these questions is to understand and use all the information in the proof of the theorem that the midpoint quadrilateral of any quadrilateral ABCD is a parallelogram. The key to that proof is that two opposite sides of the midpoint quadrilateral of ABCD are

• parallel to a diagonal
• one-half the length of that diagonal

(a) Knowing this, it is immediate that the midpoint quadrilateral is a rectangle if and only if the diagonals are perpendicular.

So then the question becomes this: Are the kites the only quadrilaterals with perpendicular diagonals? This is not hard to answer if you start with the diagonals. Just take any two segments that are perpendicular as AC and BD. To ensure that ABCD is not a kite, it is only necessary that line AC is not the perpendicular bisector of BD and line BD is not the perpendicular bisector of AC. This is true for almost all cases. For a concrete example take a point O on a segment AC or length 3, so that AO = 4 and CO = 1. Then on the line through O perpendicular to AC, choose points B and D on opposite sides of C, with OB = 2 and OD = 3.

(b) If the midpoint quadrilateral is a rhombus, all its sides are equal. But since two of the sides are one-half AC and the other two sides are one-half BD, this means ABCD is a rhombus if and only if the diagonals AC = BD.

So again start with the segments AC and BD. This time choose any two segments of equal length but which do not bisect each other. Then ABCD is not a parallelogram but the midpoint quadrilateral is a rhombus. The concrete counterexample of (a) is also a counterexample to (b).