A point O is in the interior of Δ ABC .Prove that 2(OA +OB +OC )> AB +BC +AC .
Answers
Answered by
11
Answer:
Let O be a inner point of a triangle ABC. We are to prove that
O
A
+
O
B
+
O
C
<
A
B
+
B
C
+
C
A
*Construction"
B
O
is produced to intersect AB at D.
For
Δ
A
B
D
A
B
+
A
D
>
B
D
⇒
A
B
+
A
D
>
B
O
+
O
D
...
.
.
[
1
]
For
Δ
C
O
D
C
D
+
O
D
>
O
C
...
...
[
2
]
Adding [1] and [2] we get
A
B
+
A
D
+
C
D
+
O
D
>
B
O
+
O
D
+
O
C
⇒
A
B
+
A
C
+
O
D
>
B
O
+
O
D
+
O
C
⇒
A
B
+
A
C
>
O
B
+
O
C
...
.
.
(
3
)
Similarly
A
B
+
B
C
>
O
A
+
O
C
...
.
.
(
4
)
And
A
C
+
B
C
>
O
A
+
O
B
...
.
.
(
5
)
Adding (3),(4)and (5) we get
2
(
A
C
+
B
C
+
C
A
)
>
2
(
O
A
+
O
B
+
O
C
)
⇒
O
A
+
O
B
+
O
C
<
A
B
+
B
C
+
C
A
Step-by-step explanation:
Answered by
1
Answer:
ok...............................
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