if S denotes the area of the triangle ABC, then S = area of triangles AIC+BIC+AIB but area of triangleAIC=(1/2)r*b where b denotes the length of side AC. similarly, area of triangleBIC=(1/2)r*a and area of triangleAIB=(1/2)r*c where a and c denote the lengths of sides BC and AB respectively. it is easy to note that the inradius is perpendicular to the sides as they are tangents to the incircle. thus adding the above results we get, S = (1/2)r*(a+b+c) = r*(a+b+c)/2 = r*s where s denotes the semiperimeter of triangle ABC which equals (a+b+c)/2. Q. E. D.
Drop the radii from incentre at the points of contact.. We have three kites. Area of ABC = r(s − a) + r(s − b) + r(s − c) = r(s − a + s − b + s − c) = r(3s − 2s) = rs
This is easily done by remembering that:
ReplyDeleteTr. ABC = Tr. AIC + Tr. CIB + Tr. BIA
= r*b/2 + r*a/2 + r*c/2
= r(a+b+c)/2 = r.s
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Deleteif S denotes the area of the triangle ABC, then
ReplyDeleteS = area of triangles AIC+BIC+AIB
but area of triangleAIC=(1/2)r*b where b denotes the length of side AC. similarly, area of triangleBIC=(1/2)r*a and area of triangleAIB=(1/2)r*c where a and c denote the lengths of sides BC and AB respectively. it is easy to note that the inradius is perpendicular to the sides as they are tangents to the incircle.
thus adding the above results we get,
S = (1/2)r*(a+b+c) = r*(a+b+c)/2 = r*s where s denotes the semiperimeter of triangle ABC which equals (a+b+c)/2.
Q. E. D.
Drop the radii from incentre at the points of contact.. We have three kites.
ReplyDeleteArea of ABC = r(s − a) + r(s − b) + r(s − c) = r(s − a + s − b + s − c) = r(3s − 2s) = rs
A=A1+A2+A3
ReplyDeleteA=(1/2)ar+(1/2)br+(1/2)cr
A=(1/2)r(a+b+c)
A=r.s