Question

show that bisect of angle of a parallelogram form a rectangle ​

Answer 1

Solution :

Given :

• ABCD is ||gm.
• AP, BP, CR and DR bisects Angle A, B, C and D respectively.

To prove :

• PQRS is rectangle

Proof :

In ∆ADS,

Angle DAS + Angle ADS = ½ of angle A + ½ of angle D

=> ½(angle A + angle D)

=> ½ × 180° = 90° ( co interior angles )

Again in ∆ADS, using angle sum property

Angle ADS + angle DAS + DSA = 180°

=> 90° + angle DSA = 180°

=> Angle DSA = 180° - 90°

=> Angle DSA = 90°

Also, Angle DSA = angle PSR ( vertically opposite angles are equal )

•°• Angle PSR = 90°

Similarly, Angle SPQ = angle PQR = angle QRS = 90°

•°• PQRS is a rectangle.

Hence proved!

Answer 2

★ Question :

show that bisect of angle of a parallelogram form a rectangle.

★ Answer :

let ABCD be a parallelogram.

P, Q, R and S are the points of intersection of bisectors of the angles of the parallelogram.

In ᐃ ADS

∠DAS + ADS = (1/2) ∠A + (1/2) ∠D

=( 1/2) (∠A+∠D)

=( 1/2)× 180°

= 90°

[ ∠A and ∠D are interior angles on the same side of the transversal .i.e.

∠A+∠D = 180° ]

Also,

Inᐃ ADS

DAS + ADS + DSA = 180

[ angle sum property]

⇨ 90° + ∠ DSA = 180°

∠DSA = 90°

therefore,

∠ PSR = 90°

[ vertically opposite angles ]

Similarly, it can be shown that

∠ APB or ∠ SPQ = 90°

Also,

∠ SRQ = ∠RQP = 90°

Hence, PQRS is a rectangle.