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.
