Prov the diagonale of a rectangle meet at right angles if & only if it is a square.

Given that ABCD is a rectangle and its diagonals AC and BD intersect at O such that ∠AOB = 90°.
In a rectangle, the diagonals are equal in length (AC = BD) and bisect each other (AO = OC and BO = OD).

Since AC = BD and the diagonals bisect each other, it follows that AO = OC = BO = OD.

Consider triangle AOB and triangle BOC.

• AO = CO (from Step 2)
• BO = BO (common side)
• ∠AOB = 90° (given)
• ∠BOC = 180° - ∠AOB = 180° - 90° = 90° (angles on a straight line) Therefore, triangle AOB is congruent to triangle BOC by the SAS (Side-Angle-Side) congruence criterion.

Since triangle AOB is congruent to triangle BOC, their corresponding sides are equal. Thus, AB = BC.
A rectangle with adjacent sides equal (AB = BC) is a square.

Given that ABCD is a square.
In a square, all sides are equal (AB = BC = CD = DA). A square is also a rectangle, so its diagonals are equal (AC = BD) and bisect each other (AO = OC and BO = OD).

Since AC = BD and the diagonals bisect each other, it follows that AO = OC = BO = OD.

Consider triangle AOB and triangle AOD.

• AO = AO (common side)
• BO = DO (from Step 2)
• AB = AD (sides of a square) Therefore, triangle AOB is congruent to triangle AOD by the SSS (Side-Side-Side) congruence criterion.

Since triangle AOB is congruent to triangle AOD, their corresponding angles are equal. Thus, ∠AOB = ∠AOD.
Also, B, O, D are collinear, so ∠AOB + ∠AOD = 180° (angles on a straight line).
Substituting ∠AOB for ∠AOD, we get 2 × ∠AOB = 180°.
Therefore, ∠AOB = 90°.
This means the diagonals of a square meet at right angles.

Since both directions have been proven, the diagonals of a rectangle meet at right angles if and only if it is a square.