Sides of quadrilateral are equal....prove it...!!
Answers
Explanation:
Given,
Diagonals are equal
AC=BD .......(1)
and the diagonals bisect each other at right angles
OA=OC;OB=OD ...... (2)
∠AOB= ∠BOC= ∠COD= ∠AOD= 90
0
..........(3)
Proof:
Consider △AOB and △COB
OA=OC ....[from (2)]
∠AOB= ∠COB
OB is the common side
Therefore,
△AOB≅ △COB
From SAS criteria, AB=CB
Similarly, we prove
△AOB≅ △DOA, so AB=AD
△BOC≅ △COD, so CB=DC
So, AB=AD=CB=DC ....(4)
So, in quadrilateral ABCD, both pairs of opposite sides are equal, hence ABCD is parallelogram
In △ABC and △DCB
AC=BD ...(from (1))
AB=DC ...(from $$(4)$$)
BC is the common side
△ABC≅ △DCB
So, from SSS criteria, ∠ABC= ∠DCB
Now,
AB∥CD,BC is the tansversal
∠B+∠C= 180
0
∠B+∠B= 180
0
∠B= 90
0
Hence, ABCD is a parallelogram with all sides equal and one angle is 90
0
So, ABCD is a square.
Hence proved.
Explanation:
Given,
Diagonals are equal
AC=BD .......(1)
and the diagonals bisect each other at right angles
OA=OC;OB=OD ...... (2)
∠AOB= ∠BOC= ∠COD= ∠AOD= 90
0
..........(3)
Proof:
Consider △AOB and △COB
OA=OC ....[from (2)]
∠AOB= ∠COB
OB is the common side
Therefore,
△AOB≅ △COB
From SAS criteria, AB=CB
Similarly, we prove
△AOB≅ △DOA, so AB=AD
△BOC≅ △COD, so CB=DC
So, AB=AD=CB=DC ....(4)
So, in quadrilateral ABCD, both pairs of opposite sides are equal, hence ABCD is parallelogram
In △ABC and △DCB
AC=BD ...(from (1))
AB=DC ...(from $$(4)$$)
BC is the common side
△ABC≅ △DCB
So, from SSS criteria, ∠ABC= ∠DCB
Now,
AB∥CD,BC is the tansversal
∠B+∠C= 180
0
∠B+∠B= 180
0
∠B= 90
0
Hence, ABCD is a parallelogram with all sides equal and one angle is 90
0
So, ABCD is a square.
Hence proved.