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, then the quadrilateral is a parallelogram. So, if $\overline{AC}$ and $\overline{BD}$ bisect each other at $E$, $ABCD$ is a parallelogram. In a parallelogram, opposite - sides are congruent and parallel.

Step2: Use congruent diagonals

In parallelogram $ABCD$, $\overline{AC}\cong\overline{BD}$. Consider $\triangle ABC$ and $\triangle DCB$. We know that $\overline{AB}\cong\overline{DC}$ (opposite sides of a parallelogram), $\overline{BC}\cong\overline{CB}$ (common side), and $\overline{AC}\cong\overline{BD}$. By SSS (Side - Side - Side) congruence criterion, $\triangle ABC\cong\triangle DCB$.

Step3: Prove right - angles

Since $\triangle ABC\cong\triangle DCB$, $\angle ABC\cong\angle DCB$. Also, in parallelogram $ABCD$, $\angle ABC+\angle DCB = 180^{\circ}$ (adjacent angles of a parallelogram are supplementary). So, $\angle ABC=\angle DCB = 90^{\circ}$. Since one angle of a parallelogram is a right - angle, all angles are right - angles (adjacent angles of a parallelogram are supplementary and opposite angles are equal in a parallelogram). Thus, $ABCD$ is a rectangle.

b.

1. First, draw a line segment $\overline{AC}$.
2. Using a compass, find the mid - point $E$ of $\overline{AC}$.
3. With the compass centered at $E$, draw an arc with radius $\frac{1}{2}|\overline{BD}|$ (where $|\overline{BD}| = |\overline{AC}|$).
4. Using a straightedge, draw two lines perpendicular to $\overline{AC}$ at $E$.
5. Mark the intersection points of the arc and the perpendicular lines as $B$ and $D$. Then, connect $A$, $B$, $C$, and $D$ to form the rectangle.

c.

1. Construction:

• Draw a line segment $\overline{AC}$ (the diagonal).
• Find the mid - point $E$ of $\overline{AC}$.
• Draw two lines perpendicular to $\overline{AC}$ at $E$.
• For the first rectangle in part (b), we had a certain set of distances from the mid - point of the diagonal to the vertices on the perpendiculars. For a non - congruent rectangle with the same diagonal length:
• With the compass centered at $E$, draw arcs with different radii (but such that the resulting figure is a rectangle) on the perpendicular lines to $\overline{AC}$ at $E$. Mark the intersection points as $B$ and $D$ and connect $A$, $B$, $C$, and $D$ to form a new rectangle.

1. Reason for non - congruence:

Two rectangles are congruent if and only if their corresponding side lengths are equal. Since we can vary the lengths of the sides of the rectangle while keeping the diagonal length the same (by changing the angles between the diagonals and the sides of the rectangle formed), the rectangles are not congruent. The length and width of the rectangles are different, even though the diagonal lengths are equal.

Answer:

a. Proved as above.
b. Construction steps as above.
c. Construction and reason as above.