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) why are the rectangles not congruent?
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 sss congruence, $\triangle abccong\triangle dcb$
because corresponding parts of congruent triangles are congruent, it follows that $angle abccongangle dcb$
because these angles are also interior angles on the same side of a transversal, they are supplementary.
thus, $mangle abc = mangle dcb=square^{circ}$ (simplify your answer)

Explanation:

Step1: Recall parallelogram property

Diagonals bisecting each other implies parallelogram.

Step2: Prove triangle congruence

In $\triangle ABC$ and $\triangle DCB$, $AB = DC$ (opposite - sides of parallelogram), $BC=CB$ (common side), $AC = BD$ (given). By SSS congruence, $\triangle ABC\cong\triangle DCB$.

Step3: Use congruent - triangle property

Corresponding parts of congruent triangles are congruent, so $\angle ABC\cong\angle DCB$.

Step4: Use interior - angle property

$\angle ABC$ and $\angle DCB$ are interior angles on the same side of a transversal (since $AB\parallel DC$ in parallelogram $ABCD$), so $\angle ABC+\angle DCB = 180^{\circ}$.

Step5: Solve for angle measure

Since $\angle ABC=\angle DCB$ and $\angle ABC+\angle DCB = 180^{\circ}$, then $2\angle ABC=180^{\circ}$, so $\angle ABC=\angle DCB = 90^{\circ}$. A parallelogram with one right - angle is a rectangle.

Answer:

$90$