TITLE: Proving non-collinear points of a triangle
QUESTION [0 upvotes]: My question is as follows: Let triangle $ABC$ be a triangle in $R^2$, and let $U$ be any point of $R^2$ that is not
collinear with any two of the points $A, B, C$. Let the Lines BC and AU meet
at $P, CA$ and $BU$ meet at $Q$, and $AB$ and $CU$ meet at $R$. Prove that $P, Q, R$
cannot be collinear.
I would like someone to point me in the right direction because I have no idea where to start this proof. Thank you very much!

REPLY [0 votes]: Here is a "pure geometry" approach. 
The figure formed by $A,B,C,U$ and $P,Q,R$ (left part of picture below) is very classical ; it is called a "complete quadrilateral".

I don't know if you a have a background in projective geometry ; if such is the case, you should know that this alignment property is invariant through any projective transformation, in particular any transformation $(T)$ mapping $ABCU$ onto a rectangle (as displayed on the figure) ; in this way the images $P'$ and $R'$ of $P$ and $R$ resp. belong to the line at infinity $(L)$ whereas the image $Q'$ of $Q$, which is the rectangle's center, remains at finite distance ; in particular $Q'$ does not belong to $(L)$. Thus $P',Q',R'$ aren't aligned ; therefore $P,Q,R$ aren't either.
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