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 congruence. asa sas sss aas

Explanation:

Step1: Recall parallelogram property

Since diagonals \(AC\) and \(BD\) bisect each other at \(E\), by the property of quadrilaterals, \(ABCD\) is a parallelogram with \(AB = CD\), \(AD=BC\), \(AB\parallel CD\), and \(AD\parallel BC\).

Step2: Consider triangle congruence

In \(\triangle ABE\) and \(\triangle DCE\), \(AE = DE\), \(BE = CE\) (diagonals bisect each other) and \(\angle AEB=\angle DEC\) (vertically - opposite angles). So, \(\triangle ABE\cong\triangle DCE\) (SAS congruence criterion). Similarly, \(\triangle ADE\cong\triangle BCE\). Also, since \(AC = BD\), in \(\triangle ABC\) and \(\triangle DCB\), \(AB = DC\), \(BC=CB\) and \(AC = BD\). By SSS congruence, \(\triangle ABC\cong\triangle DCB\). Then \(\angle ABC=\angle DCB\). And since \(AB\parallel CD\), \(\angle ABC+\angle DCB = 180^{\circ}\). So, \(\angle ABC=\angle DCB = 90^{\circ}\). Since one angle of the parallelogram \(ABCD\) is a right - angle, \(ABCD\) is a rectangle.

Step3: Construction of rectangle for part (b)

1. Draw a line segment \(AC\).
2. Find the mid - point \(E\) of \(AC\) using a compass and straightedge.
3. With \(E\) as the center and radius \(\frac{AC}{2}\), draw arcs above and below the line segment \(AC\).
4. Using the compass, set the width to the length of \(AC\). Draw two arcs from points \(A\) and \(C\) that intersect the previous arcs at points \(B\) and \(D\) respectively. Connect \(A\), \(B\), \(C\) and \(D\) to form the rectangle.

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

1. Construction: Draw a line segment \(AC\) of the same length as in part (b). Find its mid - point \(E\). Rotate the compass around \(E\) and mark points \(B\) and \(D\) such that \(AC\) and \(BD\) bisect each other at \(E\) and \(AC = BD\) but in a different orientation. Connect \(A\), \(B\), \(C\) and \(D\) to form a new rectangle.
2. Non - congruence: The two rectangles are not congruent because although their diagonals are congruent, the lengths of their sides can be different. The angles between the diagonals can vary, which will result in rectangles with different side - lengths while still having congruent diagonals.

Answer:

a. Proved as above.
b. Construction steps as above.
c. Construction and explanation of non - congruence as above.