October 26th, 2014, 01:27 PM  #1 
Member Joined: Nov 2010 Posts: 48 Thanks: 0  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. 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".


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case, pappus, special, theorem 
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