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

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KnowledgeBoat · Accepted Answer

In rectangle, All the interior angles equal to 90°. So, bisectors divide interior angles into two 45° angles. In △ BSC, By angle sum property of triangle, ⇒ ∠SBC + ∠SCB + ∠BSC = 180° ⇒ + ∠BSC = 180° ⇒ 45° + 45° + ∠BSC = 180° ⇒ ∠BSC = 180° - 90° = 90°. Since, ⇒ ∠SBC = ∠SCB ∴ BS = SC (Sides opposite to equal angles are equal) .....(1) From figure, ⇒ ∠PSR = ∠BSC = 90°. In △ APB and △ DRC, ⇒ ∠PAB = ∠RDC (Both equal to 45°) ⇒ ∠PBA = ∠RCD (Both equal to 45°) ⇒ AB = CD (Opposite sides of rectangle are equal) ∴ △ APB ≅ △ DRC (By A.S.A. axiom) We know that, Corresponding parts of congruent triangle are equal. ∴ BP = CR .........(2) Subtracting equation (2) from (1), we get : ⇒ BS - BP = SC - CR ⇒ PS = SR .............(3) In △ DRC, By angle sum property of triangle, ⇒ ∠RCD + ∠CDR + ∠DRC = 180° ⇒ + ∠DRC = 180° ⇒ 45° + 45° + ∠DRC = 180° ⇒ ∠DRC = 180° - 90° = 90°. From figure, ⇒ ∠SRQ = ∠DRC = 90° (Vertically opposite angles are equal) In △ PAB, By angle sum property of triangle, ⇒ ∠PAB + ∠ABP + ∠BPA = 180° ⇒ + ∠BPA = 180° ⇒ 45° + 45° + ∠BPA = 180° ⇒ ∠BPA = 180° - 90° = 90°. From figure, ⇒ ∠SPQ = ∠BPA = 90° (Vertically opposite angles are equal) In △ AQD, By angle sum property of triangle, ⇒ ∠DAQ + ∠QDA + ∠AQD = 180° ⇒ + ∠AQD = 180° ⇒ 45° + 45° + ∠AQD = 180° ⇒ ∠AQD = 180° - 90° = 90°. Since, ⇒ ∠DAQ = ∠QDA ∴ AQ = QD (Sides opposite to equal angles are equal) .....(4) From figure, ⇒ ∠PQR = ∠AQD = 90°. Since, △ APB ≅ △ DRC ∴ AP = DR .........(5) Subtracting equation (5) from (4), we get : ⇒ AQ - AP = QD - DR ⇒ PQ = QR ...........(6) In △ BSC and △ AQD, ⇒ ∠B = ∠A (Both equal to 45°) ⇒ ∠C = ∠D (Both equal to 45°) ⇒ ∠S = ∠Q (Both equal to 90°) ∴ △ BSC ≅ △ AQD (By A.A.A. axiom) We know that, Corresponding parts of congruent triangle are equal. ∴ BS = AQ .........(7) In △ AQD, ⇒ ∠A = ∠D (Both equal to 45°) ⇒ DQ = AQ (Sides opposite to equal angles are equal) .........(8) From equation (7) and (8), we get : ⇒ BS = DQ .........(9) In △ CDR, ⇒ ∠C = ∠D (Both equal to 45°) ⇒ DR = CR (Sides opposite to equal angles are equal) .........(10) From equation (2) and (10), we get : ⇒ BP = DR .....(11) Subtracting equation (11) from (9), we get : ⇒ BS - BP = DQ - DR ⇒ PS = QR .........(12) From equation (3), (6) and (12), we get : ⇒ PQ = QR = RS = PS. Since, all sides of quadrilateral PQRS are equal and each interior angle equals to 90°. ∴ PQRS is a square. Hence, proved that bisectors of the interior angles of a rectangle form a square.

Mathematics

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

Rectilinear Figures

Related Questions

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

Mathematics

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

Rectilinear Figures

Related Questions

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°.

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