If $AB$ is divided harmonically by $C,D$ and $O$ is a point not on $AB$, then prove that any transversal cuts $OA,OB,OC,OD$ in four harmonic points.
Let $OA,OB,OC,OD$ be cut by the transversal in $A',B',C',D'$ respectively. Then let $X$ be a point on $A'B'$ extended such that $ C',X$ divide $A'B'$ harmonically. So we need to prove that $X$ and $D'$ coincide. How do I proceed?
Hint:
Construct the line L that passes through AD and cut OB, OC at B’’ and C’’ respectively. Then try to prove that A, B’’, C’’, D’ is also a harmonic range. (This is an easier start because we have two lines with a common point, A.)
With respect to $\triangle ABB’’$, OC’’C and OD’D are two transversals that we can apply the Menelaus theorem.
Added:- If harmonic range (A, B, C, D) implies harmonic range (A, B'', C'', D'), then harmonic range (A, B'', C'', D') can further imply harmonic range (A', B', C', D'). (i.e. Harmonic range is transitive.)