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If $X$ and $Y$ are variable points on the sides $CA, AB$ of $\triangle ABC$ such that $CX/XA + AB/AY = 1$, prove that $XY$ passes through a fixed point

Here, using directed segments, $$CX/XA + AB/AY = 1 $$ $$CX/XA = 1 - AB/AY $$ $$CX/XA = -YB/AY$$ $$CX/XA = BY/AY$$ This results in $XY$ being a line parallel to $BC$, hence it can't possibly pass through one common point each time. Where is the mistake in this and how do I derive the answer?

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  • $\begingroup$ I don't think the equation in the question can be correct. Since lengths are positive numbers and $|AB|>|AY|$, the left hand side is bigger than $1$. $\endgroup$ – Jaap Scherphuis Apr 3 '18 at 7:00
  • $\begingroup$ @Jaap I think he means extended sides for this. $\endgroup$ – King Tut Apr 3 '18 at 7:08
  • $\begingroup$ @Jaap Directed segments mean that for a line AB, BA = -AB, in that sense the question is framed $\endgroup$ – Helix Apr 3 '18 at 7:27
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By the equation we have $\frac{CX}{XA}=\frac{YB}{AY}$ or $\frac{CX}{XA}\frac{AY}{YB}=1$. Thus by Menelaus theorem for the midpoint of $BC$, namely, $Z$, we have $\frac{CX}{XA}\frac{AY}{YB}\frac{BZ}{ZC}=1$; thus $X, Y, Z$ are collinear, as desired. (Here no directed segments are used.)

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  • $\begingroup$ Directed segments would have to be used here, so that $-YB$ can be taken into account, otherwise you won't be able to get the $-1$ required on the RHS of the Menelaus theorm. But anyway, thanks for the answer. $\endgroup$ – Helix Apr 3 '18 at 7:35

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