Euclidean Version of Pappus's theorem 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.
 A: I'm not sure what Hartshorne has in mind, but Pappus' theorem is a simple consequence of similarity of Euclidean triangles (in guise of the intercept theorem) and there's no need of introducing the circle:
Let $P$ be the point of intersection of $l$ and $m$. All we need to do is to show that $\frac{PA'}{PB} = \frac{PB'}{PA}$. 
But by assumption we have $\frac{PA'}{PC} = \frac{PC'}{PA}$ and $\frac{PC}{PB} = \frac{PB'}{PC'}$. Multiplying the left hand sides and the right hand sides together, we get what we want.
A: Without looking in too much detail, I suspect that you have not yet used the property of cyclic quadrilaterals that opposite angles are supplementary.  For example, considering the cyclic quadrilateral $ADB'C'$, $\angle B'AD$ and $\angle B'C'D$ are supplementary, so $\angle B'AD\cong\angle DC'A'$.
As a further suggestion, though I don't immediately see how to prove it, the circle through $A$, $B'$, and $C'$, the circle through $A'$, $B$, and $C'$, and the circle through $A'$, $B'$, and $C$ all pass through that same point $D$.
A: The rest of Hartshorne's hint says to see Hilbert's Foundations of Geometry (English translation currently in print by Open Court, 1971) Section III.14.  There Hilbert gives the proof using three successive cyclic quadrilaterals. 
You can find an exposition of it on-line at http://math2.uncc.edu/~frothe/3181alleuclid1_3.pdf
p. 388, where the theorem is called Pappus's theorem, following Hilbert's usage.
A: My proof is mostly by picture but using nothing outside of Euclid book I. i can fill in the details if anyone's is confused but it its pretty clear from the picture. The result regarding parallelograms is really easy to prove from propositions from book one but is common knowledge so i don't prove the result.

A: By using three cyclic quadrilaterals, we can establish the desired answer. First, let us prove an almost obvious lemma: Let ABCD be a cyclic quadrilateral, where AC is one of its diagonals. Let A'C' be parallel to AC and A' on AD and C' on extension of BC. Then A'BC'D is cyclic. The proof is quite clear: $\angle A'DB=\angle DC'B$ for ABCD is cyclic and A'C'$\parallel$AC. 
This lemma is used to show CB'A'D is cyclic, all notations as in the original thread. Hence $\alpha:=\angle CB'D=\angle DA'C$. Now we examine the following relations: $\angle BC'A'=\angle CB'C'=\angle DB'C'+\alpha =\angle DAC'+\alpha =\angle A'CD +\alpha=\angle BDA'$. They lead to the conclusion that DBA'C' is cyclic. Stacked up with the cyclic quadrilateral AB'C'D with collinear sides, the relations $\angle BA'C'=\angle ADC'=\angle PB'A$ are clear. 
This show $AB'\parallel A'B$
