Question

10 as shown in the figure, the diagonals of rectangle ( abcd ) intersect at point ( o ), ( overline{de} parallel overline{ac} ), and ( overline{ce} parallel overline{bd} ). prove that quadrilateral ( oced ) is a rhombus.

Explanation:

Step1: Prove OCED is a parallelogram

Given \( DE \parallel AC \) (i.e., \( DE \parallel OC \)) and \( CE \parallel BD \) (i.e., \( CE \parallel OD \)). By the definition of a parallelogram (a quadrilateral with both pairs of opposite sides parallel), quadrilateral \( OCED \) is a parallelogram.

Step2: Use rectangle property to show \( OC = OD \)

In rectangle \( ABCD \), the diagonals are equal and bisect each other. So, \( AC = BD \), and since \( O \) is the midpoint of \( AC \) and \( BD \), we have \( OC=\frac{1}{2}AC \) and \( OD = \frac{1}{2}BD \). Since \( AC = BD \), it follows that \( OC = OD \).

Step3: Conclude OCED is a rhombus

A parallelogram with a pair of adjacent sides equal is a rhombus. Since \( OCED \) is a parallelogram (from Step 1) and \( OC = OD \) (from Step 2), quadrilateral \( OCED \) is a rhombus.

Answer:

By proving \( OCED \) is a parallelogram (using parallel sides) and showing \( OC = OD \) (using rectangle's diagonal properties), we conclude \( OCED \) is a rhombus.