The diagonal BD of a parallelogram ABCD bisects angles B and D . Prove that ABCD is a rhombus.
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Answered by
39
Here!!
I suppose it's correct...
I suppose it's correct...
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Anshika111111111:
but square also has all its sides equal but we have to prove it a rhombus
The angles in a square are 90. But here it is not 90. So it's a rhombus.
but how do we come to know that they aren't 90
The angles in a parallelogram aren't 90. So this tells that the angles in rhombus is not 90.
OK thank you so much
Answered by
13
We have,
ABCD as the given parallelogram in
which diagonal BD bisects ∠B and ∠D.
Since, BD bisects ∠B, then
∠ABD = ∠CBD =
1
2
∠ABC ...
.(1)
Since, BD bisects ∠D, then
∠ADB = ∠CDB =
1
2
∠ADC ...
.(2)
now, ∠ABC = ∠ADC
[Opposite angles of
∥
gm
are equal]
⇒
1
2
∠ABC =
1
2
∠ADC
⇒∠ABD = ∠ADB and ∠CBD =
∠CDB [Using (1) and (2)]
⇒AD = AB and CD = BC
[Angles opposite to equal sides are equal]
.....(3)
Since, ABCD is a parallelogram, then
AB = CD and AD = BC
[Opposite sides of
∥
gm
are equal] ...
..(4)
Now, from (3) and (4), we get
AB = BC = CD = AD
⇒ABCD is a rhombus.
ABCD as the given parallelogram in
which diagonal BD bisects ∠B and ∠D.
Since, BD bisects ∠B, then
∠ABD = ∠CBD =
1
2
∠ABC ...
.(1)
Since, BD bisects ∠D, then
∠ADB = ∠CDB =
1
2
∠ADC ...
.(2)
now, ∠ABC = ∠ADC
[Opposite angles of
∥
gm
are equal]
⇒
1
2
∠ABC =
1
2
∠ADC
⇒∠ABD = ∠ADB and ∠CBD =
∠CDB [Using (1) and (2)]
⇒AD = AB and CD = BC
[Angles opposite to equal sides are equal]
.....(3)
Since, ABCD is a parallelogram, then
AB = CD and AD = BC
[Opposite sides of
∥
gm
are equal] ...
..(4)
Now, from (3) and (4), we get
AB = BC = CD = AD
⇒ABCD is a rhombus.
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