In the adjoining figure, ABCD is a parallelogram; AO and BO are the bisectors of ∠A and ∠B respectively. Prove that ∠AOB = 90°.
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✪ Given : In ||gm ABCD, AO and BO are angle bisectors of ∠A and ∠B respectively.
✪ To prove : ∠AOB = 90°
✪ Proof : Since in ||gm, opposite sides are parallel so here too AD || BC.
∠A + ∠B = 180° [Sum of interior consecutive ∠s]
Dividing above equation by 2
½ ∠A + ½ ∠B = ½ × 180°
➣ Now, ½ ∠A = ∠OAB and ½ ∠B = ∠OBA
Therefore,
∠OAB + ∠OBA = 90°
➣ In ∆AOB, due to angle sum property.
∠OAB + ∠OBA + ∠AOB = 180°
90° + ∠AOB = 180°
∠AOB = 90°
✪ Q.E.D
✪ To prove : ∠AOB = 90°
✪ Proof : Since in ||gm, opposite sides are parallel so here too AD || BC.
∠A + ∠B = 180° [Sum of interior consecutive ∠s]
Dividing above equation by 2
½ ∠A + ½ ∠B = ½ × 180°
➣ Now, ½ ∠A = ∠OAB and ½ ∠B = ∠OBA
Therefore,
∠OAB + ∠OBA = 90°
➣ In ∆AOB, due to angle sum property.
∠OAB + ∠OBA + ∠AOB = 180°
90° + ∠AOB = 180°
∠AOB = 90°
✪ Q.E.D
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