Law of Cosines/Proof 2

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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:

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:

\(\displaystyle \left({b + c}\right) \left({b - c}\right)\) \(=\) \(\displaystyle a \cdot BD\)
\(\displaystyle \implies \ \ \) \(\displaystyle \frac {b^2 - c^2} a\) \(=\) \(\displaystyle BD\)


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:

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:

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

CosineRule3.png

We extend:

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

and immediately:

$GB = CB$



$\blacksquare$