# elementary geometry/algebra

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?

• Check what you wrote. You evidently have some wrong letters in some places, because at one point you put a line parallel to $BC$ and then say it has an endpoint on $BC$. You probably meant $AC$ in the second instance; what other wrong letters are there? – David K Mar 15 '15 at 4:19
• I fixed the one you pointed out. I'm sorry. I'm very tired at this point. – user223709 Mar 15 '15 at 4:20
• Also fixed two more. AY/YB was supposed to be AY/YC – user223709 Mar 15 '15 at 4:21
• OK. If you just need to know about the ratio at the end then the rest is just an introduction. – David K Mar 15 '15 at 4:21

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 $\dfrac xy = \dfrac ab.$

• Thank you. I guess my algebraic manipulation involving fractions is a bit weak. – user223709 Mar 15 '15 at 4:26

If $\frac{x}{x+y} = \frac{a}{a+b}$ then it implies $\frac{a+b}{a} = \frac{x+y}{x} \implies 1 + \frac{b}{a} = 1 + \frac{y}{x} \implies \frac{x}{y} = \frac{a}{b}$