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

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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