# 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.

-

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.

-
Thank you for your response Theo. I have seen the proof using Thales' theorem, and indeed it is much nicer! I believe Hartshorne refrains from using this as his book only assumes knowledge of the first four books of the Elements thus far. I would like to use only methods from those books, as proportions haven't been introduced. Nevertheless, thank you. – yunone Jan 31 '11 at 22:45
@yunone: Ah, I see. Since I don't know what the first four books of Euclid cover (shame on me), I guess I can't help you... – t.b. Jan 31 '11 at 22:58

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 $D$.

-
Thank you Isaac, I will try to investigate further. – yunone Jan 31 '11 at 20:20
So $\angle B'AC'\cong\angle B'DC'\cong\gamma$. Since $\angle ADC'$ and $\angle AB'C'$ are supplementary, $\angle AB'C'\cong\angle BDC'$. If we assume that $DBA'C'$ is a cyclic quadrilateral, then $\angle BA'C'\cong\angle ADC'$ as they are both supplementary to $\angle BDC'$. But $\angle ADB'\cong\angle AC'B'\cong\angle CA'C'$, and thus subtracting , we have $\angle B'AC'\cong\angle BA'C$, showing that $AB'\parallel BA'$. I guess it all hinges on showing $DBA'C'$ is cyclic, but I don't see how to show that either. One lemma I know is that it will be cyclic if... – yunone Jan 31 '11 at 20:34
...$\angle C'DA'\cong\angle C'BA'$. – yunone Jan 31 '11 at 20:35