Question

Prove that a parallelogram is a rectangle if it's diagonals are congruent

Answer 1

To prove that a parallelogram is a rectangle if its diagonals are congruent, we can follow these steps:

1. **Definition of a Parallelogram**: A parallelogram is a quadrilateral with opposite sides that are parallel.

2. **Properties of Parallelograms**: In a parallelogram, opposite sides are equal in length, and opposite angles are equal. Additionally, the diagonals bisect each other.

3. **Congruent Diagonals**: Suppose we have a parallelogram \( ABCD \) with diagonals \( AC \) and \( BD \) that are congruent, i.e., \( AC = BD \).

4. **Midpoints of the Diagonals**: Let \( O \) be the point of intersection of the diagonals \( AC \) and \( BD \). Since the diagonals of a parallelogram bisect each other, we have:
\[
AO = OC \quad \text{and} \quad BO = OD
\]

5. **Expressing the Lengths**: Let \( AO = OC = m \) and \( BO = OD = n \). By the property of the diagonals being congruent, we have:
\[
AC = AO + OC = m + m = 2m
\]
\[
BD = BO + OD = n + n = 2n
\]
Setting these equal because the diagonals are congruent:
\[
2m = 2n
\]
This simplifies to:
\[
m = n
\]

6. **Right Triangles**: Now, consider triangle \( AOB \) and triangle \( COD \). Since \( AO = OC = m \) and \( BO = OD = m \), both triangles have two equal sides.

7. **Angle Relationships**: Angles \( AOB \) and \( COD \) are vertically opposite angles, hence they are equal. Therefore, by the Side-Angle-Side (SAS) criterion for triangle congruence, \( \triangle AOB \cong \triangle COD \).

8. **Implication of Triangle Congruence**: Since the triangles are congruent, all corresponding sides are equal:
\[
AB = CD \quad \text{and} \quad AD = BC
\]
This confirms that opposite sides are equal, which is already established in a parallelogram.

9. **Right Angles in the Parallelogram**: Next, observe that because the diagonal \( AC \) equals \( BD \), triangles \( AOB \) and \( COD \) can be considered further for right angles:
Since each triangle \( AOB \) and \( BOC \) congruently holds \( AO = OB \) and \( CO = OD \), we can infer that the angles \( AOB \) and \( COD \) are \( 90^\circ \), leading to:
\[
\angle AOB = \angle BOC = 90^\circ
\]

10. **Conclusion**: Since all angles in quadrilateral \( ABCD \) formed by diagonal intersections are \( 90^\circ \), \( ABCD \) can be concluded to be a rectangle. Therefore we conclude:

\[
\text{A parallelogram is a rectangle if its diagonals are congruent.}
\]

This completes the proof.