Given triangle ABC with side lengths a, b, c and A', B', C' are points on sidelines BC, CA, AB respectively. (O) is circumcircle of ABC with radius R. A'', B'', C'' are intesections of AA', BB', CC' with (O) respectively (other than A, B, C).
We define six signed circumradii of six triangles:
r(BA') = circumradius of triangle BA''A'
r(CA') = circumradius of triangle CA''A'
r(CB') = circumradius of triangle CB''B'
r(AB') = circumradius of triangle AB''B'
r(AC') = circumradius of triangle AC''C'
r(BC') = circumradius of triangle BC''C'
If vertices of the triangle follow the directions of vertices of ABC then its circumradius is positive, otherwise its circumradius is negative. Similarly we define signed segments on the sidelines of ABC. Of course r(XY) = - r(YX). From now, all radii and segments are signed.

Angle(AA''B) = Angle(ACB) or = 180 - Angle(ACB)
therefore (by sinus theorem):
r(BA')/R = BA'/AB = BA'/c and r(BA') = (BA'/c)*R
Similarly: r(CA') = (CA'/b)*R or r(A'C) = (A'C/b)*R
From these: r(BA')/r(A'C) = (BA'/A'C)*(b/c)
Similarly we can show:
r(BA')/r(A'C) = (BA'/A'C)*(b/c)
r(CB')/r(B'A) = (CB'/B'A)*(c/a)
r(AC')/r(C'B) = (AC'/C'B)*(a/b)
Multiply these three expressions we have:
r(BA')/r(A'C)*r(CB')/r(B'A)*r(AC')/r(C'B) =
(BA'/A'C)*(CB'/B'A)*(AC'/C'B)
The right part is part of Ceva theorem so we can say:

r(BA')/r(A'C)*r(CB')/r(B'A)*r(AC')/r(C'B) = 1 if and only if AA', BB', CC' are concurrent.

Because this fact related Ceva theorem so I am not sure that this fact is well known or not. Please kindly give me your reference if you know about it'somewhere!

Thank you and best regards,
Bui Quang Tuan