Dear dolly

given

(a-b)/(b-c)=(s-a)/(s-c)

1/(s-a)(b-c)  = 1/(s-c)(a-b)

Δ/(s-a)(b-c)  = Δ/(s-c)(a-b)

r1/(b-c)  = r3/(a-b)  .................1

now  we know   (b-c)/r1   + (c-a)/r2   + (a-b)/r3  =0

    so from equation 1     2(b-c)/r1  = -(c-a)/r2

                                             r1/(b-c)  = -2r2/(c-a)

 so r1 : r2: r3  :: (b-c):(a-c)/2 : (a-b)

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