Question

Prove that quadrilateral formed by the intersection of angle bisectors of all angles of a parallelogram is a rectangle. (Figure 5.24)




Ans btaiye.

Answer 1

Hello Mate!

Given : ABCD is ||gm whose angle bisectors meet to form EFGH.

To prove : EFGH is rectangle

Proof : < BAD + < ADC = 180°

½ <BAD + ½ <ADC = 90°

< DAH + < ADH = 90°

In ∆ ADH

< DAH + < ADH + < AHD = 180°

90° + < AHD = 180°

< AHD = 90°

< AHD = < GHE = 90° ( Vertically opp. angles ]

Similarly, < HGF = < GFE = < HEF = 90°

Hence, all angles are 90° so EFGH is rectangle.

Hence proved

Have great future ahead!

Answer 2

Let QWER be a parallelogram 

To prove Quadrilateral LMNO is a rectangle.

Hence :-

QWER is a parallelogram, then RE || QW and RQ is a transversal.

∠Q +∠R= 180°

⇒ 1/2 ∠M +  1/2 ∠N= 90° [dividing both sides by 2]

∠OQR + ∠ORQ = 90°

∠QOR = 90°   

∴ ∠LON = 90° and ∠LMN = 90°     [vertically opposite angles]

∠MNO = 90°and ∠MLO = 90° [V●O●A]

So, LMNO is a quadrilateral whose each angle is 90°.

Hence, LMNO is a rectangle