Constructing a point in a projective line given other seven points and two cross-ratios

Question

Let's suppose we have eight points $A,B,C,X,A',B',C',X'$ on the same projective line such that $(ABCX) = (A'B'C'X')$. If the first seven points are known, how could I find the eighth point $X'$?

Answer 1

Name the given line $L'$. Introduce a point $P$ not on $L'$ and a different line $L''$ not containing $P$ nor any of the other given points.

Define $A'',B'',C'',X''$ as points of intersection of the lines $\overline{AP},\overline{BP},\overline{CP},\overline{XP}$ with $L''$. By construction, these new points have the same cross-ratio as $A,B,C,X$.

Then use this thread to construct $X'$ on $L'$ such that $X''=A''$ would result in $X'=A'$ etc and that $A',B',C',X'$ have the same cross-ratio as $A'',B'',C'',X''$.