Triangle Sine and Circle Radius

To round up my sines and cosines, in an earlier post I forgot to mention that I have to elaborate a bit about the law of Sines. If you draw a circle through the three vertices of the triangle this is called the circumscribed circle. If the circumscribed circle has a diameter of 2r the Law of Sines states that

    \[\frac{a}{\sin\alpha}=\frac{b}{\sin\beta}=\frac{c}{\sin\gamma}=2r\]

You can prove this as follows (As always: click on the pictures for a larger version):

You see a diagonal line running through the circle with a length (of course) of 2r. This line both splits the big triangle upper angle \angle ACB and the small orange isosceles triangle upper angle \angle AOB below. But there are also two isosceles triangles to the left and the right with angles h (left) and i (right). Three angles make \pi radians. Thus for the remaining angle of both triangles in the middle (\angle AOC and \angle BOC) it means

    \[\pi-2h\]

and

    \[\pi-2h\]

Now look at the remaining angles of those two with the diagonal of the circle. That total angle also adds up to \pi. So the remaining two triangles, which both add up to the top angle of the orange isosceles triangle are respectively

    \[\pi-(\pi-2h)=2h\]

and

    \[\pi-(\pi-2i)=2i\]

This means that for the two triangles that share the same circle sector of the circumscribed circle the top angles we can write:
if the top angle of the large triangle is h+i=\alpha then the top angle of the small triangle is 2h+2i=2(h+i) or \angle AOB=2\cdot\angle ACB.

So we can write those angles as \alpha and 2\alpha:

In this picture the orange isosceles triangle below thus has a top angle of 2\alpha. You see the interior bisector of the top angle. From this it follows that both the angles to the right and the left of the bisector ar \frac{2\alpha}{2}=\alpha. The basis of the triangle with length a is also divided in two, so each half has length \tfrac{1}{2}a.
Now we can calculate sin\alpha:

    \[\sin\aplha=\frac{\tfrac{1}{2}a}{r}=\frac{a}{2r}\]

Calculating \frac{a}{\sin\alpha} we get:

    \[\frac{a}{\sin\alpha}=\frac{a}{\frac{a}{2r}}=\frac{2r\cdot a}{a}=2r\]

and thus

    \[\frac{a}{\sin\alpha}=\frac{b}{\sin\beta}=\frac{c}{\sin\gamma}=2r\]