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October 26th, 2014, 01:27 PM   #1
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Pappus' Theorem Special Case

I'm trying to solve Pappus' theorem from a book of plane geometry. The theorem states that given one set of collinear points $\displaystyle P,Q,R$ and another set of collinear points $\displaystyle P',Q',R'$ then the intersection points $\displaystyle X={P'Q} \cap {Q'P},Y={PR'} \cap {P'R},Z={QR'} \cap {Q'R}$ are collinear.
The book solves the first case which is when $\displaystyle P'Q \nparallel Q'R$ , then we apply Menelaus' theorem at the triangle$\displaystyle \bigtriangleup ABC$ where $\displaystyle A={P'Q} \cap {PR'},B={P'Q} \cap {Q'R},C={Q'R} \cap {R'P}$ and at the lines $\displaystyle QR',P'R,Q'P,PQ,P'Q'$.
The second case is when $\displaystyle P'Q \parallel RQ'$. Here I follow like the first case and apply Menelaus' theorem at the triangle $\displaystyle \bigtriangleup ABC$ where $\displaystyle A={PQ'} \cap {P'R},B={PQ'} \cap {QR'},C={QR'} \cap {RP'}$ and at the lines $\displaystyle R'P,P'Q,Q'R, PQ,P'Q'$.
The third case and the one that troubles me more is when $\displaystyle P'Q \parallel Q'R,PQ' \parallel QR'$. A hint the problem gives is to prove that the lines $\displaystyle PP'$ and $\displaystyle QQ'$ are parallel and use similar triangles.
At first I consider the case where the lines passing through $\displaystyle P,Q,R$ and $\displaystyle P',Q',R'$ are not parallels and meet at a point $\displaystyle O$. Because the lines $\displaystyle P'Q$ and $\displaystyle Q'R$ are parallel we have $\displaystyle \frac{OQ}{OR}=\frac{OP'}{OQ'}$, also because the lines $\displaystyle PQ'$ and $\displaystyle QR'$ are parallel we have $\displaystyle \frac{OP}{OQ}=\frac{OQ'}{OR'}$. Multiplying these two relations we get $\displaystyle \frac{OP}{OR}=\frac{OP'}{OR'}$ which means that $\displaystyle PP'$ and $\displaystyle QQ'$ are parallel.
A pair of similar triangle is $\displaystyle \bigtriangleup PYP'$ and $\displaystyle \bigtriangleup R'YR$. With this informtion (I don't know if there are other similar triangles) I'm trying to find a triangle and use the Menelaus' theorem to prove that $\displaystyle X,Y,Z$ are colinear. I tried much but didn't make it. Any idea?
I know there are proofs using projective geometry but I need a proof only in the context of plane geometry.
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Last edited by talisman; October 26th, 2014 at 01:30 PM. October 27th, 2014, 01:05 PM #2 Member   Joined: Nov 2010 Posts: 48 Thanks: 0 I didn't mean "$\displaystyle PP'$ and $\displaystyle QQ'$ are parallels" but "$\displaystyle PP'$ and $\displaystyle RR'$ are parallels". Tags case, pappus, special, theorem Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Leen Number Theory 5 April 11th, 2014 11:51 AM maxgeo Algebra 1 October 31st, 2012 03:57 PM Numerator Calculus 1 January 11th, 2010 06:46 PM dionysos Real Analysis 0 August 27th, 2009 08:40 AM

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