Bisectors of the Angles of a Parallelogram form a Rectangle

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°.









9th Grade Math

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