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

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

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 (Side - Side - Side) congruence criterion, $\triangle ABC\cong\triangle DCB$.

Step3: Use corresponding - parts of congruent triangles

Since $\triangle ABC\cong\triangle DCB$, $\angle ABC\cong\angle DCB$ (corresponding parts of congruent triangles are congruent).

Step4: Use interior - angle property

$\angle ABC$ and $\angle DCB$ are interior angles on the same side of the transversal $BC$ with $AB\parallel DC$. For parallel lines $AB$ and $DC$ cut by transversal $BC$, $\angle ABC+\angle DCB = 180^{\circ}$.

Step5: Solve for angle measure

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

Answer:

$90$