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- #1

- Apr 14, 2013

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Let $ABC$ be a triangle and let $A'\in BC$, $B'\in AC$ and $C'\in AB$. I want to show the following:

If $A',B',C'\neq A,B,C$ and either zero or two of the points $A',B',C'$ are at one side of triangle, then $A',B',C'$ are collinear iff $\frac{|AC'|}{|C'B|}=\frac{|BA'|}{|A'C|}=\frac{|CB'|}{|B'A|}=1$.

Could you give me a hint how we could show that?