## Friday, November 28, 2014

### 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.