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. abcd is a kite and its adjacent sides are congruent. abcd is a rhombus and all of its opposite sides are congruent. abcd is a trapezoid

Explanation:

Step1: Recall parallelogram property

If the diagonals of a quadrilateral bisect each other, by the definition of parallelogram (a quadrilateral is a parallelogram if its diagonals bisect each other), $ABCD$ is a parallelogram and its opposite sides are congruent and parallel.

Let the diagonals $AC$ and $BD$ bisect each other at point $E$. So $AE = EC$ and $BE=ED$. Also, given $AC = BD$, then $AE=EC = BE = ED$.

Step2: Consider triangles formed by diagonals

In $\triangle ABE$ and $\triangle DCE$, we have $AE = EC$, $\angle AEB=\angle DEC$ (vertically - opposite angles) and $BE = ED$. So, $\triangle ABE\cong\triangle DCE$ (by the Side - Angle - Side (SAS) congruence criterion). Similarly, $\triangle ADE\cong\triangle BCE$.

In $\triangle ABC$ and $\triangle DCB$, $AB = DC$ (opposite sides of parallelogram), $BC=CB$ (common side) and $AC = BD$ (given). So, $\triangle ABC\cong\triangle DCB$ (by the Side - Side - Side (SSS) congruence criterion). Then $\angle ABC=\angle DCB$.

Since $AB\parallel DC$ in parallelogram $ABCD$, $\angle ABC+\angle DCB = 180^{\circ}$ (adjacent angles of a parallelogram are supplementary). Since $\angle ABC=\angle DCB$, we have $\angle ABC=\angle DCB = 90^{\circ}$.

Since a parallelogram with one right - angle is a rectangle, $ABCD$ is a rectangle.

Step3: Construction in part (b)

1. Draw a line segment $AC$.
2. Construct the perpendicular bisector of $AC$. Let the mid - point of $AC$ be $E$.
3. With $E$ as the center and radius equal to $\frac{1}{2}AC$, use a compass to draw arcs above and below the perpendicular bisector of $AC$.
4. Let the intersection points of the arcs and the perpendicular bisector be $B$ and $D$. Connect $AB$, $BC$, $CD$ and $DA$. The quadrilateral $ABCD$ is a rectangle by part (a) because the diagonals $AC$ and $BD$ (where $BD = AC$) bisect each other.

Step4: Construction and non - congruence in part (c)

1. Draw a line segment $AC$ of the desired length for the diagonal.
2. Construct the perpendicular bisector of $AC$. Let the mid - point be $E$.
3. On the perpendicular bisector of $AC$, choose different points $B$ and $B'$ (and corresponding $D$ and $D'$ such that $E$ is the mid - point of $BD$ and $B'D'$ and $BD=B'D' = AC$).
4. The rectangles are not congruent because although the diagonals are of the same length, the lengths of the sides of the rectangles can be different. In a rectangle, if the diagonal is $d$ and the sides are $a$ and $b$, by the Pythagorean theorem $d^{2}=a^{2}+b^{2}$. For different pairs of non - negative real numbers $(a,b)$ such that $d^{2}=a^{2}+b^{2}$, we get non - congruent rectangles.

Answer:

a. See the above - step by - step proof that a quadrilateral with congruent and bisecting diagonals is a rectangle.
b. See the above - described construction steps using a compass and straightedge.
c. See the above - described construction steps for a non - congruent rectangle with the same diagonal length. The rectangles are not congruent because the side - lengths of the rectangles can be different while the diagonals have the same length.