Here we will prove that the bisectors of the angles of a parallelogram form a rectangle.
Given: PQRS is a parallelogram in which PQ ∥ SR and SP ∥ RQ. The bisectors of ∠P, ∠Q, ∠R and ∠S are PJ, QK, RL and SM respectively which enclose the quadrilateral JKLM.
To prove: JKLM is a rectangle.
Proof:
Statement |
Reason |
1. ∠QPS + ∠PSR = 180° Therefore, \(\frac{1}{2}\)∠QPS + \(\frac{1}{2}\)∠PSR = 90° |
1. PQ ∥ SR. |
2. ∠SPM + ∠PSM = 90° |
2. PJ and SM are bisectors of ∠QPS and ∠PSR respectively. |
3. ∠PMS = 90° ⟹ JM ⊥ ML. |
3. Sum of the three angles of ∆PSM is 180°. |
4. Taking bisectors of ∠S and ∠R, ML ⊥ LK; Taking bisectors of ∠R and ∠Q, KL ⊥ JK; Taking bisectors of ∠Q and ∠P, JK ⊥ JM. |
4. Similarly. |
5. JK ∥ ML, JM ∥ KL. |
5. Two lines perpendicular to the same line are parallel. |
6. JKLM is a parallelogram. (Proved). |
6. By statement 5 and one angle, say ∠JML = 90°. |
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