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.

A Rhombus is a Parallelogram whose Diagonals Meet at Right Angles


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.






9th Grade Math

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