Prove that the bisectors of interior angles of a parallelogram form a rectangle.

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KnowledgeBoat · Accepted Answer

Let ABCD be the parallelogram. From figure, AH, BE, CF and DG are bisectors of ∠A, ∠B, ∠C and ∠D respectively. We know that, Consecutive angles of parallelogram are supplementary. Considering ∠A + ∠B = 180°, ⇒ ⇒ = 90° ⇒ = 90° ............(1) Considering ∠B + ∠C = 180°, ⇒ ⇒ = 90° ⇒ = 90° ............(2) Considering ∠C + ∠D = 180°, ⇒ ⇒ = 90° ⇒ = 90° .........(3) Considering ∠D + ∠A = 180°, ⇒ ⇒ = 90° ⇒ = 90° .........(4) In △ PAB, By angle sum property of triangle, ⇒ ∠PAB + ∠ABP + ∠BPA = 180° ⇒ + ∠BPA = 180° ⇒ 90° + ∠BPA = 180° [From equation (1)] ⇒ ∠BPA = 180° - 90° = 90°. From figure, ⇒ ∠SPQ = ∠BPA = 90° (Vertically opposite angles are equal) In △ BSC, By angle sum property of triangle, ⇒ ∠SBC + ∠SCB + ∠BSC = 180° ⇒ + ∠BSC = 180° ⇒ 90° + ∠BSC = 180° [From equation (2)] ⇒ ∠BSC = 180° - 90° = 90°. From figure, ⇒ ∠PSR = ∠BSC = 90°. In △ DRC, By angle sum property of triangle, ⇒ ∠RCD + ∠CDR + ∠DRC = 180° ⇒ + ∠DRC = 180° ⇒ 90° + ∠DRC = 180° [From equation (3)] ⇒ ∠DRC = 180° - 90° = 90°. From figure, ⇒ ∠SRQ = ∠DRC = 90° (Vertically opposite angles are equal) In △ AQD, By angle sum property of triangle, ⇒ ∠QAD + ∠ADQ + ∠DQA = 180° ⇒ + ∠DQA = 180° ⇒ 90° + ∠DQA = 180° [From equation (4)] ⇒ ∠DQA = 180° - 90° = 90°. From figure, ⇒ ∠RQP = ∠DQA = 90°. Since, all the interior angles of quadrilateral PQRS equals to 90°. ∴ PQRS is a rectangle. Hence, proved that bisectors of interior angles of a parallelogram form a rectangle.

Mathematics

Prove that the bisectors of interior angles of a parallelogram form a rectangle.

Rectilinear Figures

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In parallelogram ABCD, the bisectors of angle A meets DC at P and AB = 2AD.
Prove that :
(i) BP bisects angle B.
(ii) Angle APB = 90°.

Mathematics

Prove that the bisectors of interior angles of a parallelogram form a rectangle.

Rectilinear Figures

Related Questions

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In parallelogram ABCD, the bisectors of angle A meets DC at P and AB = 2AD.Prove that :(i) BP bisects angle B.(ii) Angle APB = 90°.

In parallelogram ABCD, the bisectors of angle A meets DC at P and AB = 2AD.
Prove that :
(i) BP bisects angle B.
(ii) Angle APB = 90°.