An homographic transformation transforms three points $A$, $B$, $C$ of a line into the points $A'$, $B'$, $C'$. Prove that it transforms the bound points of the homography that exchanges $A$, $B$, $C$ cyclically into the the bound points of homography that exchanges $A'$, $B'$, $C'$ cyclically.
I'm studying homographies and I found this exercises, but I don't understand very well how to begin it. Any help would be appreciate.
Let $\pi$ be the projectivity that sends $(A,B,C)$ to $(A',B',C')$ and $\tau$ be the projectivity that sends $(A,B,C)$ to $(B,C,A)$. Let $X$ be a fixed point of $\tau$. Then $$\pi\circ\tau\circ\pi^{-1}$$ sends $(A',B',C',\pi(X))$ to $(B',C',A',\pi(X))$. It means that $\pi(X)$ is a fixed point of the projectivity that sends $(A',B',C')$ to $(B',C',A')$, as we had to prove.