# Law of Cosines/Proof 2

## Contents

## Theorem

Let $\triangle ABC$ be a triangle whose sides $a, b, c$ are such that $a$ is opposite $A$, $b$ is opposite $B$ and $c$ is opposite $C$.

Then:

- $c^2 = a^2 + b^2 - 2 a b \cos C$

## Proof

Let $\triangle ABC$ be a triangle.

### Case 1: $AC$ greater than $AB$

Using $AC$ as the radius, we construct a circle whose center is $A$.

Now we extend:

- $CB$ to $D$
- $AB$ to $F$
- $BA$ to $G$
- $CA$ to $E$.

$D$ is joined with $E$, thus:

Using the Intersecting Chord Theorem we have:

- $GB \cdot BF = CB \cdot BD$

$AF$ is a radius, so $AF = AC = b = GA$ and thus:

- $GB = GA + AB = b + c$
- $BF = AF - AB = b - c$

Thus:

\(\displaystyle \left({b + c}\right) \left({b - c}\right)\) | \(=\) | \(\displaystyle a \cdot BD\) | |||||||||||

\(\displaystyle \implies \ \ \) | \(\displaystyle \frac {b^2 - c^2} a\) | \(=\) | \(\displaystyle BD\) |

Next:

\(\displaystyle CD\) | \(=\) | \(\displaystyle CB + BD\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle a + \frac {b^2 - c^2} a\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \frac {a^2 + b^2 - c^2} a\) |

As $CA$ is a radius, $CE$ is a diameter.

By Thales' Theorem, it follows that $\angle CDE$ is a right angle.

Then using the definition of cosine, we have

\(\displaystyle \cos C\) | \(=\) | \(\displaystyle \frac {CD} {CE}\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \frac {\left({\dfrac{a^2 + b^2 - c^2} a}\right)} {2 b}\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \frac {a^2 + b^2 - c^2} {2 a b}\) | |||||||||||

\(\displaystyle \implies \ \ \) | \(\displaystyle c^2\) | \(=\) | \(\displaystyle a^2 + b^2 - 2 a b \cos C\) |

$\Box$

### Case 2: $AC$ less than $AB$

When $AC$ is less than $AB$, the point $B$ lies outside the circle and so the diagram needs to be modified accordingly:

Now we extend:

- $BA$ to $G$
- $CA$ to $E$.

Then we construct:

- $D$ as the point at which $CB$ intersects the circle
- $F$ as the point at which $AB$ intersects the circle.

Finally $D$ is joined to $E$.

Using the Secant Secant Theorem we have:

- $GB \cdot BF = CB \cdot BD$

$AF$ is a radius, so $AF = AC = b = GA$ and thus:

- $GB = GA + AB = b + c$
- $BF = AB - AF = b - c$

Thus:

\(\displaystyle \left({b + c}\right) \left({b - c}\right)\) | \(=\) | \(\displaystyle CB \cdot BD\) | Secant Secant Theorem | ||||||||||

\(\displaystyle \left({b + c}\right) \left({b - c}\right)\) | \(=\) | \(\displaystyle a \cdot BD\) | |||||||||||

\(\displaystyle \implies \ \ \) | \(\displaystyle \frac {b^2 - c^2} a\) | \(=\) | \(\displaystyle BD\) |

Next:

\(\displaystyle CD\) | \(=\) | \(\displaystyle CB - BD\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle a - \frac {b^2 - c^2} a\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \frac {a^2 - b^2 + c^2} a\) |

As $CA$ is a radius, $CE$ is a diameter.

By Thales' Theorem, it follows that $\angle CDE$ is a right angle.

Then using the definition of cosine, we have

\(\displaystyle \cos C\) | \(=\) | \(\displaystyle \frac {CD} {CE}\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \frac {\left({\dfrac {a^2 - b^2 + c^2} a}\right)} {2 b}\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \frac {a^2 - b^2 + c^2} {2 a b}\) | |||||||||||

\(\displaystyle \implies \ \ \) | \(\displaystyle c^2\) | \(=\) | \(\displaystyle a^2 + b^2 - 2 a b \cos C\) |

$\Box$

### Case 3: $AC = AB$

When $AC = AB$ the points $B$, $D$ and $F$ coincide on the circumference of the circle:

We extend:

- $BA$ to $G$
- $CA$ to $E$

and immediately:

- $GB = CB$

$\blacksquare$