A Rhombus is a Parallelogram whose Diagonals Meet at Right Angles

Here we will prove that a rhombus is a parallelogram whose diagonals meet at right angles.

Given: PQRS is a rhombus. So, by definition,

PQ = QR = RD = SP. Its diagonals PR and QS intersect at O.

To prove: (i) PQRS is a parallelogram.

(ii) ∠POQ = ∠QOR = ∠ROS = ∠SOP = 90°.

Proof:

Statement	Reason
(i) In ∆PQR and ∆RSP, 1. PQ = RS and QR = PS	1. Given.
2. PR = RP	2. Common side
3. ∆PQR ≅ ∆RSP Therefore, ∠QPR = ∠SRP, ∠QRP = ∠SPR.	3. By SSS criterion of congruency. CPCTC
4. SR ∥ PQ, PS ∥QR.	4. Alternate angles are equal.
5. PQRS is a parallelogram. (Proved) (ii) In ∆OPQ and ∆ORS,	5. By definition.
6. ∠OPQ = ∠ORS	6. By statement 4, PQ ∥ SR and PR is a transversal.
7. ∠OQP = ∠OSR	7. P PQ ∥ SR and QS is a transversal
8. PQ = SR	8. Given.
9. ∆OPQ ≅ ∆ORS Therefore, OP = OR, OQ= OS. In ∆POS ≅ ∆ROS,	9. By AAS criterion of congruency. CPCTC
10. PS = RS	10. Given.
11. OP = OR	11. From statement 10.
12. OS = SO	12. Common side.
13. Therefore, ∆POS ≅ ∆ROS	13. By SSS criterion of congruency.
14. ∠POS = ∠ROS	14. CPCTC
15. ∠POS + ∠ROS = 180°	15. Linear pair.
16. ∠POS = ∠ROS = 90°	16. From statements 14 and 15.
17. ∠POQ = ∠ROS, ∠QOR = ∠POS Therefore, ∠POQ = ∠QOR =∠ROS = ∠SOP = 90° (Proved)	17. Opposite angles.

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