Proof of the sine rule

So I made my first attempt at a proof. I think it turned out well. Maybe not. But I was wondering if someone could take a look at it and tell me what they think. I'd be glad to hear some criticism on it. So here it goes:

The Sine Rule is as follows: $$\frac{\sin(\alpha)}{a}=\frac{\sin(\beta)}{b}=\frac{\sin(\gamma)}{c}$$ $\mathbf{Proof}$: We are given an acute triangle $\triangle OPQ$ with sides $a$,$b$, and $c$. Opposite of the sides are the angles $\alpha$,$\beta$, and $\gamma$, respectively. We will divide $\triangle OPQ$ with $2$ line segments, $h_1$ and $h_2$. The line segment $h_1$ will have an endpoint at angle $\gamma$ and extend to side $c$ to make $h_1$ perpendicular to side $c$. The line segment $h_2$ will have an endpoint at angle $\alpha$ and will extend to side $a$ to make $h_2$ perpendicular to side $a$. We know that $\sin(\alpha)=\frac{h_1}{b}$ and $\sin(\beta)=\frac{h_1}{a}$. We can deduce that $h_1=a\sin(\beta)$ and $h_1=b\sin(\alpha)$. From these two equations, we can produce the following result: \begin{align}a\sin(\beta)=b\sin(\alpha)\\ \frac{\sin(\beta)}{b}=\frac{\sin(\alpha)}{a}\end{align} We now turn our attention to see that $\sin(\gamma)=\frac{h_2}{b}$ and $\sin(\beta)=\frac{h_2}{c}$. As before we can deduce that $h_2=b\sin(\gamma)$ and $h_2=c\sin(\beta)$. From this system of equations we can simplify it in the following way: \begin{align}b\sin(\gamma)=c\sin(\beta)\\ \frac{\sin(\gamma)}{c}=\frac{\sin(\beta)}{b}\end{align} It follows that if $\frac{\sin(\alpha)}{a}=\frac{\sin(\beta)}{b}$, and $\frac{\sin(\beta)}{b}=\frac{\sin(\gamma)}{c}$, then $\frac{\sin(\alpha)}{a}=\frac{\sin(\gamma)}{c}$. Thus proving that $$\frac{\sin(\alpha)}{a}=\frac{\sin(\beta)}{b}=\frac{\sin(\gamma)}{c}$$ Which concludes my proof. I know it's a pretty basic concept learned in high school trig but I just wanted to start off on something easy. ANY input would be great.

• One minor suggestion: you want to prove the sine rule and you state it before the proof. However you don't say what $a, b$ and $c$ are until the beginning of the proof. So the statement just above the proof is actually not a statement since the reader doesn't know what $a,b$ and $c$ are. Was I clear? So my suggestion is: let the reader know what $a, b$ and $c$ before you start the proof. – Git Gud Jan 20 '13 at 21:13
• When you get $\frac{\sin(\alpha)}{a} = \frac{\sin(\beta)}{b}$, it's not really useful to restart in the same way to get the same thing with $b$ and $c$. We generally say "the problem being symmetrical (in $a$, $b$, $c$), we have the same equality for other sides". – Nicolas Jan 20 '13 at 22:33
• Looks fine, you need to see what happens when your triangle isn't acute as well, though. Right-angle is trivial, so just take a look at obtuse one. – Lazar Ljubenović Apr 6 '13 at 9:16

$$\sin\left(\alpha\right)=\frac{H_c}{b}$$ $$\sin\left(\beta\right)=\frac{H_c}{a}$$ $$H_c=b\sin(\alpha)$$ $$H_c=a\sin(\beta)$$ $$a\sin(\beta)=b\sin(\alpha)$$ $$\frac{a}{\sin(\alpha)}=\frac{b}{\sin(\beta)}$$
• Could you explain why it's not necessary to demonstrate that these are equal to $c / \sin(C)$ as well please? – baxx May 13 at 9:58