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