**Law of Sines Proof**

The Law of Sines is a/(sin A) = b/(sin B) = c/(sin C) = the diameter of the
circumscribed circle.

Draw an altitude of the triangle (extending side c, if necessary):

From the diagram, it is clear that h=a sin B, and that h= b sin A. (If
side c needed to be extended, because angle B is larger than 90 degrees, then
note that the sine of B is equal to the sine of the supplement of B, so even
though h is to the right of B, it is still true that h=a sin B.)

So a sin B = b sin A,

so a/(sin A) = b/(sin B)

By drawing a different altitude, the same can be shown of any two sides, proving
the first part of it, namely that

a/(sin A) = b/(sin B) = c/(sin C).Now for the second part, that c/(sin C) is
equal to the diameter of the circumscribed circle. Consider angle C, which
is subtended by chord c of the circle:

Angle C is the same as angle D in the diagram, below, because it is subtended
by the same chord. (Or, perhaps angle D is supplementary to angle C, which
happens if C is obtuse).

Since triangle ABD is a right triangle, c/(sin D) is the length of AD, the
diameter of the circle. Since the measure of C either equals the measure
of D or is supplementary to D, their sines are equal, so c/(sin D) = c/(sin C) =
the diameter of the circle. This proves the second part.

### Related pages in this website:

Inscribed Angle Property
-- that all angles that are inscribed in a circle that are subtended by a
given chord have equal measure, and that measure is half the central angle
subtended by the same chord.

Circumscribed Circle
- The radius of a circle circumscribed around a triangle is R = abc/(4K),
where K is the area of the triangle.

Laws of Sines and Cosines - and using
these laws to find the missing sides and angles in a triangle

The webmaster and author of this Math Help site is
Graeme McRae.