Law of Cosines - Proofs - Using Geometry of The Circle

Using Geometry of The Circle

Using the geometry of the circle, it is possible to give a more geometric proof than using the Pythagorean theorem alone. Algebraic manipulations (in particular the binomial theorem) are avoided.

Case of acute angle γ, where a > 2b cos γ. Drop the perpendicular from A onto a = BC, creating a line segment of length b cos γ. Duplicate the right triangle to form the isosceles triangle ACP. Construct the circle with center A and radius b, and its tangent h = BH through B. The tangent h forms a right angle with the radius b (Euclid's Elements: Book 3, Proposition 18; or see here), so the yellow triangle in Figure 8 is right. Apply the Pythagorean theorem to obtain

Then use the tangent secant theorem (Euclid's Elements: Book 3, Proposition 36), which says that the square on the tangent through a point B outside the circle is equal to the product of the two lines segments (from B) created by any secant of the circle through B. In the present case: BH2 = BC BP, or

Substituting into the previous equation gives the law of cosines:

Note that h2 is the power of the point B with respect to the circle. The use of the Pythagorean theorem and the tangent secant theorem can be replaced by a single application of the power of a point theorem.

Case of acute angle γ, where a < 2b cos γ. Drop the perpendicular from A onto a = BC, creating a line segment of length b cos γ. Duplicate the right triangle to form the isosceles triangle ACP. Construct the circle with center A and radius b, and a chord through B perpendicular to c = AB, half of which is h = BH. Apply the Pythagorean theorem to obtain

Now use the chord theorem (Euclid's Elements: Book 3, Proposition 35), which says that if two chords intersect, the product of the two line segments obtained on one chord is equal to the product of the two line segments obtained on the other chord. In the present case: BH2 = BC BP, or

Substituting into the previous equation gives the law of cosines:

Note that the power of the point B with respect to the circle has the negative value −h2.

Case of obtuse angle γ. This proof uses the power of a point theorem directly, without the auxiliary triangles obtained by constructing a tangent or a chord. Construct a circle with center B and radius a (see Figure 9), which intersects the secant through A and C in C and K. The power of the point A with respect to the circle is equal to both AB2 − BC2 and AC·AK. Therefore,

begin{align} c^2 - a^2 & {} = b(b + 2acos(pi - gamma)) \ & {} = b(b - 2acosgamma), end{align}

which is the law of cosines.

Using algebraic measures for line segments (allowing negative numbers as lengths of segments) the case of obtuse angle (CK > 0) and acute angle (CK < 0) can be treated simultaneously.