In triangle ABC, prove that rri . =(s-b)(s-c)
Answers
Answered by
1
solution:
Given a+b+c=2s
a+b=2s−c
a+c=2s−b
We know that
cosA=
2bc
b
2
+c
2
−2a
2
2sin
2
(
2
A
)=1−cosA
=1−(
2bc
b
2
+c
2
−a
2
)
=
2bc
a
2
+2bc−(b
2
+c
2
)
2bc
a
2
−(b
2
+c
2
−2bc)
=
2bc
a
2
−(b−c)
2
=
2bc
(a+b−c)(a−b+c)
=
2bc
(2s−c−c)(2s−b−b)
=
2bc
2(s−c)2(s−b)
bc
2(s−b)(s−c)
2sin
2
2
A
=2.
bc
(s−b)(s−c)
2sin
2
2
A
=
bc
(s−b)(s−c)
sin(
2
A
)=
bc
(s−b)(s−c)
Hence proved.
Similar questions