Derivation of r=(s-a)tan(A/2) = (s-b)tan(B/2) = (s-c)tan(C/2)

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Derivation of

 r = (s-a)tan(A/2) = (s-b)tan(B/2)

 = (s-c)tan(C/2) 

Where ,

 r = inradius of  triangle ABC

 s = semi perimeter of triangle ABC

 a,b and c are the lengths of sides BC,AC, and   AB respectively of   triangle ABC.            

Before going to the derivation, let us learn some formulae required for this derivation.

Required formulae :

In above triangle ABC,

tan(A/2) = ∆/(s(s-a)) , 

tan(B/2) = ∆/(s(s-b)) ,

tan(C/2) = ∆/(s(s-c)) ,

r = ∆/s

Let us go into derivation.

From above above formulae,

tan(A/2) = ∆/(s(s-a)) implies 

∆/s = (s-a)tan(A/2) 

similarly,

tan(B/2) = ∆/(s(s-b))  implies 

∆/s  = (s-b)tan(B/2) and

tan(C/2) = ∆/(s(s-c))  implies  ∆/s = (s-c)tan(C/2)

and r = ∆/s

From above information,

r = ∆/s = (s-a)tan(A/2)  = (s-b)tan(B/2) = (s-c)tan(C/2) 

Therefore, r = (s-a)tan(A/2) = (s-b)tan(B/2) = (s-c)tan(C/2) 



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