# 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°.

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