in a rhombus ABCD, prove that 2AB>BD
Answers
Answer:
Step-by-step explanation:same length.)
A
B
=
B
C
=
C
D
=
D
A
=
a
2) Opposite angles of a rhombus are congruent (the same size and measure.)
∠
B
A
D
=
∠
B
C
D
=
y
,
and
∠
A
B
C
=
∠
A
D
C
=
x
3) The intersection of the diagonals of a rhombus form 90 degree (right) angles. This means that they are perpendicular.
∠
A
O
B
=
∠
B
O
C
=
∠
C
O
D
=
∠
D
O
A
=
90
∘
4) The diagonals of a rhombus bisect each other. This means that they cut each other in half.
B
O
=
O
D
=
1
2
B
D
=
m
,
and
A
O
=
O
C
=
1
2
A
C
=
n
5) Adjacent sides of a rhombus are supplementary. This means that their measures add up to 180 degrees.
x
+
y
=
180
∘
Now back to our question.
In
Δ
B
O
C
,
B
C
2
=
B
O
2
+
O
C
2
Since
B
O
=
1
2
B
D
,
and
O
C
=
1
2
A
C
⇒
B
C
2
=
(
1
2
B
D
)
2
+
(
1
2
A
C
)
2
⇒
B
C
2
=
1
4
(
B
D
)
2
+
1
4
(
A
C
)
2
⇒
B
C
2
=
1
4
(
B
D
2
+
A
C
2
)
⇒
4
B
C
2
=
A
C
2
+
B
D
2
Answer:
let the point of intersection is O. Take Rt anglr triangle ABO.
BD= 2BO
In the rhombus , half BD is the side of the Rt angle triangle of ABO.
& AB is the hypotenuse.
Since hypotenuse is always MORE the the sides of Rt angle Triangle, The given problem is proved.
Step-by-step explanation: