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?
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.)