Question

a. prove that a quadrilateral whose diagonals are congruent and bisect each other is a rectangle.
b. explain how to use part (a) and only a compass and straightedge to construct any rectangle.
c. construct another rectangle not congruent to the rectangle in part (b) but whose diagonals are congruent to the diagonals of the rectangle in part (b).
a. let $overline{ac}$ and $overline{bd}$ be two line segments that bisect each other at e, with $overline{ac}congoverline{bd}$. prove that abcd is a rectangle.
because $overline{ac}$ and $overline{bd}$ bisect each other, abcd is a parallelogram and its opposite sides are congruent and parallel.
due to congruence.

Explanation:

Step1: Recall parallelogram property

Since diagonals $\overline{AC}$ and $\overline{BD}$ bisect each other at $E$, quadrilateral $ABCD$ is a parallelogram with $AB = CD$, $AD=BC$, $AB\parallel CD$ and $AD\parallel BC$.

Step2: Consider triangle congruence

In $\triangle AEB$ and $\triangle DEC$, $AE = EC$, $BE=ED$ (diagonals bisect each other) and $\angle AEB=\angle DEC$ (vertically - opposite angles). Also, since $AC = BD$, then $AE = EC=BE = ED$. In $\triangle AEB$ and $\triangle AED$, $AE = AE$, $BE = ED$ and $AB = AD$ (sides of parallelogram). By SSS congruence criterion, $\triangle AEB\cong\triangle AED$.

Step3: Prove right - angles

Since $\triangle AEB\cong\triangle AED$, $\angle AEB+\angle AED = 180^{\circ}$ and $\angle AEB=\angle AED = 90^{\circ}$. Since adjacent angles in a parallelogram are supplementary and one angle is a right - angle, all angles of the parallelogram $ABCD$ are right - angles. So, $ABCD$ is a rectangle.

Answer:

SAS is not the correct congruence criterion here. The correct congruence criterion is SSS. Because we know that $AE = EC$, $BE = ED$ (diagonals bisect each other) and $AC = BD$ implies $AE=EC = BE=ED$. Then in triangles formed by the diagonals and sides of the parallelogram, we have three pairs of equal sides. So the answer for the first blank is SSS.