Question

show that if the diagonals of a quadrilateral are equal and bisect each other at right angles to write on a page in short

Answer 1

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.

Answer 2

Step-by-step explanation:

Given that ABCD is a square.

To prove : AC=BD and AC and BD bisect each other at right angles.

Proof:

(i) In a ΔABC and ΔBAD,

AB=AB ( common line)

BC=AD ( opppsite sides of a square)

∠ABC=∠BAD ( = 90° )

ΔABC≅ΔBAD( By SAS property)

AC=BD ( by CPCT).

(ii) In a ΔOAD and ΔOCB,

AD=CB ( opposite sides of a square)

∠OAD=∠OCB ( transversal AC )

∠ODA=∠OBC ( transversal BD )

ΔOAD≅ΔOCB (ASA property)

OA=OC ---------(i)

Similarly OB=OD ----------(ii)

From (i) and (ii) AC and BD bisect each other.

Now in a ΔOBA and ΔODA,

OB=OD ( from (ii) )

BA=DA

OA=OA ( common line )

ΔAOB=ΔAOD----(iii) ( by CPCT

∠AOB+∠AOD=180° (linear pair)

2∠AOB=180°

∠AOB=∠AOD=90°

∴AC and BD bisect each other at right angles.