I was looking at my geometry chapter summary on similar triangles, and I was a little confused with the result. I'm really tired right now and I am having difficulty leafing through the chapter to confirm it. I tried to prove it myself. Anyway, here is what's confusing me. $\bigtriangleup ABC$ has a line segment $\overline{XY}$ parallel to $\overline{BC}$ going though $\bigtriangleup ABC$ with endpoints on sides $AB$ and $AC$. Since triangles ABC and AXY are similar, it is clear that $AX/AB=AY/AC$.
But the book asserts that $AX/XB=AY/YC$. This wasn't very clear to me, as $AX/XB=AY/YC$ aren't really similarity statements because they aren't ratios between corresponding sides, so I found an algebraic translation. So now I'm left to prove this seemingly simple statement: If $\frac{x}{x+y}=\frac{a}{a+b}$, then $\frac{x}{y}=\frac{a}{b}$. I can't reduce it any further. Maybe my book was wrong?
If $\dfrac{x}{x+y}=\dfrac{a}{a+b}$, then $\dfrac{x+y}{x} = \dfrac{a+b}{a}$, so $$\frac yx = \frac{x+y}{x} - 1 = \frac{a+b}{a} - 1 = \frac ba.$$ Therefore $\dfrac xy = \dfrac ab.$