(A) /AB’I = 2/ACI = C(1)(angles subtended by arc AI) /CB’I = 2/CAI = A(angles subtended by arc CI)

Adding:/AB’C = C + A = 180^{o} – B, so the angles AB’C and
ABC are supplementary, making the quadrilateral AB’CB cyclic,
and proving that B’ lies on the circumcircle of triangle ABC.
By similar reasoning, A’ and C’ also lie on the circumcircle.

Using (1) in the isosceles triangle IAB’ gives

/AIB’ = 90^{o}
– C/2 = A/2 + B/2

But A/2 and B/2 are internal angles of triangle ABI, so /AIB’
must be its external angle, proving that B, I, B’ are collinear,
and that BI crosses the circumcircle at B’.
By similar reasoning, AI and CI cross the circumcircle at A’ & C’

(B)Let O denote the circumcentre of
triangles ABC and A’B’C’.

/A’OC’ = 2 /A’B’C’(angles subtended by arc A’C’) = 2(/A’B’B
+ /C’B’B) = 2(/A’AB +
/C’CB) = 2(A/2 + C/2) = 180^{o}
– B Area of triangle A’OC’ = [A’OC’] = (1/2)R^{2 }sin(180 – B)

=
(1/2)R^{2} sin B

and by similar reasoning: Area of triangle A’B’C’= [B’OC’] + [C’OA’] + [A’OB’]

=
(1/2)R^{2}(sinA +sinB + sinC)

= R(2RsinA
+2RsinB +2RsinC)/4

= R(a
+ b + c)/4

Since area of triangle ABC = r(a + b +
c)/2, it follows that