Geometrically showing $\frac{\alpha}{\beta} > \frac{\sin\alpha}{\sin\beta}$, for $0 < \beta < \alpha < 90^\circ$ 
I attempting to prove the show trig identity, i.e. $\frac{\alpha}{\beta} > \frac{\sin\alpha}{\sin\beta}$. I have found that the radius is $1$ so there are three line segments equal to $1$, but I am unsure which segment would be $\frac{\sin\alpha}{\sin\beta}$?

$$\sin x < x < \tan x \qquad \forall~x \in [0, \frac{\pi}2]$$
If one agrees to start with the above known fact (which is purely an intuitive geometric fact as detail in e.g. this answer and NOT about algebraic analysis of functions), then:
On one hand, we have 
$$ \frac{\alpha - \beta}2 > \sin \bigl( \frac{ \alpha -\beta}2\bigr) \quad \implies\quad \alpha - \beta > 2 \sin \bigl( \frac{ \alpha -\beta}2\bigr)  \tag*{Eq.(1)}$$
On the other hand, $ \alpha > \beta$ and cosine being a decreasing function yield
$$ \beta < \tan\beta = \frac{ \sin\beta }{ \cos\beta } < \frac{ \sin\beta }{ \cos\bigl( \frac{ \alpha + \beta}2 \bigr) } \quad \implies \quad \sin\beta > \beta \cdot \cos\bigl( \frac{ \alpha + \beta}2 \bigr)  \tag*{Eq.(2)}
$$
The product of the two inequalities (all terms are positive) gives us
\begin{align}
&& (\alpha - \beta) \cdot \sin\beta &> \beta \cdot 2 \sin\bigl( \frac{ \alpha - \beta}2 \bigr)\cos\bigl( \frac{ \alpha + \beta}2 \bigr) \\
&\implies & (\alpha - \beta) \cdot \sin\beta &> \beta \cdot (\sin\alpha - \sin\beta) \\
&\implies & \frac{\alpha - \beta}{ \beta } &> \frac{\sin\alpha - \sin\beta}{ \sin\beta}\qquad \text{, then $+1$ on both sides} \\
&\implies & \frac{ \alpha }{ \beta } &> \frac{ \sin\alpha }{ \sin\beta}
\end{align}
Note that the whole derivation can be viewed as a purely geometric proof expressed via algebra. One can make a sketch and find ALL the relevant lengths (including e.g. $\frac{ \sin\beta }{ \cos( \frac{ \alpha + \beta}2 ) }$).
