I'm going through Hartshorne's Geometry, and one of the exercises has stumped me for a good few hours. The problem is a version of one of Pappus's theorems:
Let $A$, $B$, $C$, be points on a line $l$, and let $A'$, $B'$, $C'$ be points on a line $m$. Assume $AC'\parallel A'C$ and $B'C\parallel BC'$. Show that $AB'\parallel A'B$.
A hint is given to draw a circle through $A$, $B'$, $C'$ meeting $l$ in $D$, and to use cyclic quadrilaterals.
I've included a picture for clarity:
To the right, I've drawn lines $DB'$ and $DC'$. From the cyclic quadrilateral, $ADB'C'$, I can see that $\angle B'AC'\cong\angle B'DC'$, as they are on the same circumference. I call this angle $\gamma$. Similarly, $\angle C'B'D\cong\angle C'AD$. This allowed me to prove that $\triangle AEC,\triangle CFB,\triangle B'C'E,\triangle C'A'F$ are all similar. I attempted to show $\triangle AEB'$ and $\triangle A'FB$ are similar, but was unsuccessful.
My strategy was to show that $\angle BA'C=\gamma$, and this would suffice to show $AB'\parallel BA'$ since $AC'\parallel CA'$, but I haven't quite managed it. Perhaps someone sees the way to finish off this theorem? Thank you for your help.