We know the following result from Trigonometry,
IO=R√1−8sinA2sinB2sinC2 ..... (1)
and r=4RsinA2sinB2sinC2 ..... (2)
$(IO)^2 \, = \, R^2 \, - \, 8R^2 \, \sin \dfrac{A}{2} \, \sin \dfrac{B}{2} \, \sin
\dfrac{C}{2}$
= R2 - 2R(r) = R2 - 2Rr, by (1) and (2).
2nd part :
For Δ BIO to be right angled triangled, we have
BI2+IO2=BO2
From part (i) above
BI=4RsinA2sinC2and(IO)2=R2−2Rr
∴16R2sin2(A/2)sin2(C/2)+R2−2R.(4Rsin(A/2)sin(B/2)sin(C/2))=R2
or 2sin(A/2)sin(C/2)=sin(B/2))=R2
Now putting the values of \sin (A/2) etc . in terms of sides . we get
2(s−b)b=1 or 2s=3b or a+c=2b
i.e.b is A.M. of a and c.