Question

show that if the diagonals of a quadrilateral are equal and bisect each other at right angle then it is a square also created by short

Answer 1

thanks mark as BRAINLIST

Answer 2

We have a quadrilateral ABCD such that ��O�� is the mid-point of AC and BD. Also AC ⊥ BD.

           Now, in ΔAOD and ΔAOB, we have

                 AO = AO

[Common]

                 OD = OB

[∵ O is the mid-point of BD]

                 ∠AOD = ∠AOB

[Each = 90°]

                 ∴ ΔAOD ≌ ∠AOB

[SAS criteria]

                 ∴Their corresponding parts are equal.

                 ⇒ AD = AB

...(1)

           Similarly, we have

AB = BC

...(2)

 

BC = CD

...(3)

 

CD = DA

...(4)

           From (1), (2), (3) and (4) we have: AB = BC = CD = DA

           ∴Quadrilateral ABCD is having all sides equal.

           In ΔAOD and ΔCOB, we have

 

AO = CO

[Given]

 

OD = OB

[Given]

 

∠AOD = ∠COB

[Vertically opposite angles]

           ∴

ΔAOD ≌ ΔCOB

 

           ⇒Their corresponding pacts are equal.

           ⇒                       ∠1 = ∠2

           But, they form a pair of interior alternate angles.

           ∴AD || BC

           Similarly, AB || DC

           ∴ ABCD is' a parallelogram.

           ∵ Parallelogram having all of its sides equal is a rhombus.

           ∴ ABCD is a rhombus.

           Now, in ΔABC and ΔBAD, we have

 

AC = BD

[Given]

 

BC = AD

[Proved]

 

AB = BA

[Common]

 

ΔABC ≌ ΔBAD

[SSS criteria]

 

           Their corresponding angles are equal.

 

∠ABC = ∠BAD

 

           Since, AD || BC and AB is a transversal.

           ∴∠ABC + ∠BAD = 180°

[Interior opposite angles are supplementary]

           i.e. The rhombus ABCD is having one angle equal to 90°.

           Thus, ABCD is a square.