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Proof of Pappus' Theorem

Date: 04/02/2001 at 14:12:56
From: Kornelius Boersma
Subject: Pappus' Theorem

How can I prove Pappus' theorem of collinearity with the help of 
Menelaus' theorem? I've tried many ways, but I couldn't get it.

Date: 04/03/2001 at 05:12:05
From: Doctor Floor
Subject: Re: Pappus' Theorem

Hi, Kornelius,

Thanks for writing.

We have a hexagon ABCDEF inscribed in a pair of lines (i.e. they are 
alternating on one line and the other). Let P be the point of 
intersection of opposite sides AB and DE, Q of BC and EF, and R of CD 
and FA. Then Pappus' theorem states that PQR are collinear.

To prove this with the help of Menelaus' theorem, we use triangle XYZ, 
formed by the sides AB, CD and EF. Here's the figure:

Applying Menelaus' theorem to BC, DE and AF we get:

     BX   QY   CZ
     -- * -- * -- = -1   ..........[1]
     BY   QZ   CX

     PX   EY   DZ
     -- * -- * -- = -1   ..........[2]
     PY   EZ   DX

     AX   FY   RZ
     -- * -- * -- = -1   ..........[3]
     AY   FZ   RX

while applying Menelaus' theorem to lines ACE and BDF we find:

     AX   EY   CZ
     -- * -- * -- = -1   ..........[4]
     AY   EZ   CX

     BX   FY   DZ
     -- * -- * -- = -1   ..........[5]
     BY   FZ   DX

Now if we multiply the left-hand sides of [1], [2], [3] and divide by 
the product of the left-hand sides of [4], [5], and we do the same for 
the right-hand sides of these equations, we find:

     PX   QY   RZ
     -- * -- * -- = -1
     PY   QZ   RX

By (the converse of) Menelaus' theorem this gives that PQR are 

If you need more help, just write back.

Best regards,
- Doctor Floor, The Math Forum   
Associated Topics:
High School Geometry
High School Triangles and Other Polygons

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