>AA* = 2*Rab*Rac/R (1)
>Similarly we can have
>BB* = 2*Rbc*Rba/R (2)
>CC* = 2*Rca*Rcb/R (3)
>Do the same with triangle A'B'C' we can have:
>A'A'' = 2*Rca*Rba/R (1')
>B'B'' = 2*Rab*Rcb/R (2')
>C'C'' = 2*Rbc*Rac/R (3')

Dear Greg and Alex,

From my proof we can show one interesting fact for any inscribed hexagon (not necessary convex) AB'CA'BC'.
From (1), (2), (3), (1'), (2'), (3')
k*l*m = p*q*r = 8*Rab*Rac*Rbc*Rba*Rca*Rca =
8*(AB'/2)*(AC'/2)*(BC'/2)*(BA'/2)*(CA'/2)*(CB'/2)/R^3
If we denote six sides of inscribed hexagon AB'CA'BC' as a, b, c, d, e, f then

k*l*m = p*q*r = a*b*c*d*e*f/(2*R)^3 = a*b*c*d*e*f / W^3 or

a*b*c*d*e*f = k*l*m*W^3 = p*q*r*W^3 (4)
where
W = diameter of circumcircle of the hexagon
p, q, r = distances from A, B, C to sidelines of A'B'C'
k, l, m = distances from A', B', C' to sidelines of ABC

If can we somehow generalize (4) for other inscribed polygon?

Best regards,
Bui Quang Tuan