Dear Greg and Alex,

I have found one following proof, please check for me if I make some mistakes.
Let A, B, C, A', B', C' are concyclic on one circle (O), radius R
Let A'', B'', C'' are orthogonal projection of A'B'C' on BC, CA, AB respectively.
Let A*, B*, C* are orthogonal projection of ABC on B'C', C'A', A'B' respectively.

We can easy show some following concyclic point sets:
A, A*, B', B'' are concyclic on circle (Ab) (center Ab = midpoint of AB'), radius = Rab
A, A*, C', C'' are concyclic on circle (Ac) (center Ac = midpoint of AC'), radius = Rac
B, B*, C', C'' are concyclic on circle (Bc) (center Bc = midpoint of BC'), radius = Rbc
B, B*, A', A'' are concyclic on circle (Ba) (center Ba = midpoint of BA'), radius = Rba
C, C*, A', A'' are concyclic on circle (Ca) (center Ca = midpoint of CA'), radius = Rca
C, C*, B', B'' are concyclic on circle (Cb) (center Cb = midpoint of CB'), radius = Rcb
Moreover:
A, Ac, Ab, O are concyclic on circle (Oa), (center Oa = midpoint of OA), radius R/2
B, Bc, Ba, O are concyclic on circle (Ob), (center Ob = midpoint of OB), radius R/2
C, Ca, Cb, O are concyclic on circle (Oc), (center Oc = midpoint of OC), radius R/2
By some similar triangles we can also easy show following fact:
The common chord d of two given circles with radius R1, R2 can be calculated as:
d = R1*R2/R3 where R3 = radius of the circle passing through two centers of given circle and one their intersection point.
We now consider three circles at vertex A: (Ac), (Ab), (Oa), we can show:
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')
From (1), (2), (3), (1'), (2'), (3') we have the result:
AA* x BB* x CC* = A'A'' x B'B'' x C'C'' or
k*l*m = p*q*r

Best regards,
Bui Quang Tuan