Inscribed Triangles and the Law of Sines

The Law of Sines follows from the following fact:  

Theorem: Any side of a triangle divided by the sine of the opposite angle gives the diameter of the circumscribing circle.

In other words, $\frac{a}{\sin \alpha}, \frac{b}{\sin \beta}$, and $\frac{c}{\sin \gamma}$ are all equal because each represents the same diameter.

The Law of Sines can be proven by using the fact that the area of a triangle is half the product of any two sides and the sine of the included angle. But such a proof gives no hint of the geometric interpretation of $\frac{a}{\sin \alpha}$.

Here's a geometric argument for the theorem.

Case 1: First, we notice that if we have a right triangle with hypotenuse $c$, then $\sin\alpha =\frac{a}{c}$, $\sin\beta =\frac{b}{c}$, and $\sin\gamma =\sin 90^{\circ}=1$, so all three ratios  $\frac{a}{\sin \alpha}, \frac{b}{\sin \beta}$, and $\frac{c}{\sin \gamma}$  are equal to $c$, the hypotenuse. And because an inscribed right angle subtends an arc of $180^{\circ}$, the hypotenuse coincides with a diameter. Thus the theorem is true for all angles in any right triangle.

Case 2: Now suppose that $\Delta ABC$ is a triangle with $\alpha =\angle CAB$ an acute angle. Draw the diameter through $B$ to the point $D$, and draw the segment from $D$ to $C$.

$\Delta DBC$ is a right triangle inscribed in the same circle and shares the side of length $a$ with $\Delta ABC$. The angle at $D$ subtends the same arc as the angle at $A$, so the angles are congruent. By Case 1, $\frac{a}{\sin\alpha}$ is the diameter of the circumscribing circle. Thus the theorem holds for any acute angle in any triangle.

Case 3: Now suppose that  $\Delta ABC$ is a triangle with $\alpha =\angle CAB$ an obtuse angle. Choose $D$ so it lies on the arc of the circle subtended by $\angle CAB$. Then $\Delta DBC$ is inscribed in the same circle as $\Delta ABC$ and shares the side of length $a$.

Because $\angle CAB$ and $\angle CDB$ subtend arcs that sum to $360^{\circ}$, they are supplementary angles. In particular, $\angle CDB$ is acute, so the diameter of the circumscribing circle is $\frac{a}{\sin(180^{\circ}-\alpha)}$. And because $\sin(180^{\circ}-\alpha) = \sin\alpha$, we conclude that the diameter of the circumscribing circle is  $\frac{a}{\sin \alpha}$.

Thus, whether $\alpha$ is a right angle, acute angle, or obtuse angle, the ratio  $\frac{a}{\sin \alpha}$ gives the diameter of the circumscribing circle.