Law of Cosines/Proof 2

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:


 * CosineRule.png

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:

Next:

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

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:


 * CosineRule2.png

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:

Next:

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

Case 3: $AC = AB$
When $AC = AB$ the points $B$, $D$ and $F$ coincide on the circumference of the circle:


 * CosineRule3.png

We extend:
 * $BA$ to $G$
 * $CA$ to $E$

and immediately:
 * $GB = CB$