Proof of Pappus' TheoremDate: 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 collinear. If you need more help, just write back. Best regards, - Doctor Floor, The Math Forum http://mathforum.org/dr.math/ |
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