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°
⇒ $\frac{∠B}{2} + \frac{∠C}{2}$ + ∠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°
⇒ $\frac{∠C}{2} + \frac{∠D}{2}$ + ∠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°
⇒ $\frac{∠A}{2} + \frac{∠B}{2}$ + ∠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°
⇒ $\frac{∠A}{2} + \frac{∠D}{2}$ + ∠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.
