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.