Question

show that the bisector of the angle of a parallelogram form a rectangle​

Answer 1

Answer:

Given Let ABCD be a parallelogram and AP, BR, CR, be are the bisectors of ∠A, ∠B, ∠C and ∠D, respectively.

To prove Quadrilateral PQRS is a rectangle.

Proof Since, ABCD is a parallelogram, then DC || AB and DA is a transversal.

∠A+∠D= 180°

[sum of cointerior angles of a parallelogram is 180°]

⇒ 1/2 ∠A+  1/2 ∠D = 90° [dividing both sides by 2]

∠PAD + ∠PDA = 90°

∠APD = 90°    [since,sum of all angles of a triangle is 180°]

∴ ∠SPQ = 90°     [vertically opposite angles]

∠PQR = 90°

∠QRS = 90°

and ∠PSR = 90°

Thus, PQRS is a quadrilateral whose each angle is 90°.

Hence, PQRS is a rectangle.

Answer 2

Answer:

Given : A parallelogram ABCD.

To prove : PQRS is a rectangle.

Proof :

In ΔABS ,

1/2∠A + 1/2∠B + ∠BSA = 180 °

∠A+∠B = 180 ° (Adjacent angles of a parallelogram are supplementary )

∴ 1/2∠A + 1/2∠B = 180 ÷ 2 = 90 °

90 ° + ∠BSA = 180 °

∴ ∠BSA = 180° - 90°

∠BSA = 90°

∠BSA = ∠RSP [Vertically opposite angles ]

∠RSP = 90 °

Similarly , it can be showed that ∠SPQ = 90° , ∠PQR = 90° and ∠QRS = 90°

∴ PQRS is a quadrilateral in which all the angles are of measure 90°.

Hence , PQRS is a rectangle