Ex 8.1, 4
Show that the diagonals of a square are equal and bisect each other at right angles.
Given: ABCD be a square.
Diagonals intersect at O.
To prove : We need to prove 3 things
1. The diagonals of a square are equal ,i.e. AC = BD
2. bisect each other, i.e. OA = OC & OB = OD,
3. at right angles ,any of AOB , BOC , COD , AOD is 90
Proof:
In ABC and DCB,
AB = DC
ABC = DCB
BC = BC
ABC DCB
AC = DB
Hence, the diagonals of a square are equal in length.
Now we need to prove diagonals bisect each other i.e. AO = CO, BO = DO
In AOB and COD,
AOB = COD
ABO = CDO
AB = CD
AOB COD
AO = CO and OB = OD
Hence, the diagonals of a square bisect each other.
In AOB and COB,
OA = OC
AB = BC
BO = BO
AOB COB
AOB = COB
Now
AOB + COB = 180
AOB + AOB = 180
2 AOB = 180
AOB = (180 )/2
AOB = 90
Hence, AC & BD bisect at right angles
Hence proved
Ex 8.1, 4 (Method 2)
Show that the diagonals of a square are equal and bisect each other at right angles.
Ex 8.1, 4 (Method 2)
Show that the diagonals of a square are equal and bisect each other at right angles.
Given: ABCD be a square.
Diagonals intersect at O.
To prove: We need to prove 3 things
The diagonals of a square are equal ,i.e. AC = BD
Bisect each other, i.e. OA = OC & OB = OD
At right angles, any of ∠AOB , ∠BOC , ∠COD , ∠AOD is 90°
Proof:
In ΔABC and ΔDCB,
AB = DC
∠ABC = ∠DCB
BC = CB
∴ ΔABC ≅ ΔDCB
Therefore
AC = DB
Hence, the diagonals of a square are equal in length
(Sides of square are equal)
(Both 90° , as all angles of square are 90°)
(Common)
( SAS congruence rule)
( CPCT)
…(1)
Now we need to prove diagonals bisect each other
i.e. AO = CO, BO = DO
Since Square is a parallelogram
& Diagonals of a parallelogram bisect each other
Hence, the diagonals of a square bisect each other.
i.e., OA = OC
& OB = OD
Now, we have to prove diagonals bisect each other at right angles
In ΔAOB and ΔCOB,
OA = OC
AB = CB
BO = BO
∴ ΔAOB ≅ ΔCOB
Therefore
∠AOB = ∠COB
Now
∠AOB + ∠COB = 180°
∠AOB + ∠AOB = 180°
2∠AOB = 180°
(From (2))
(Sides of square are equal)
(Common)
(SSS congruency)
(CPCT)
(Linear Pair)
(From (3))
∠AOB = (180° )/2
∠AOB = 90°
Therefore, AC & BD bisect at right angles
Hence proved

Article by

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.