# 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 ≅ ∆RSPTherefore, ∠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 ≅ ∆ORSTherefore, 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 = ∠POSTherefore, ∠POQ = ∠QOR =∠ROS = ∠SOP = 90° (Proved) 17. Opposite angles.

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