Yet Another Proof of the Law of Cosines

The law of cosines relates the lengths of the sides of a triangle to the cosine of one of its angles. Using standard notation, the law of cosines states

$$c^2=a^2+b^2-2ab\cos{\gamma},$$

where $\gamma$ denotes the angle contained between sides of lengths $a$ and $b$ and opposite the side of length $c$. For the same figure, the other two relations are analogous:

$$a^2=b^2+c^2-2ac\cos{\alpha},$$

$$b^2=a^2+c^2-2ac\cos{\beta}.$$

Proof. Let $D$, $E$ and $F$ be the contact points of the incircle with $AC$, $AB$ and $BC$, respectively. Also, let $AE=AD=x$; $BE=BF=y$; $CD=CF=z$. We start from two well-known relationships of a triangle: $$\sin^2{\frac{\gamma}{2}}=\frac{(s-a)(s-b)}{ab} \qquad\text{and}\qquad \cos^2{\frac{\gamma}{2}}=\frac{s(s-c)}{ab}$$  

(See Cut-the-knot's Relations between various elements of a triangle for proofs), where $s$ denotes the semiperimeter of $\triangle{ABC}$. Since $(s-a)=x$, $(s-b)=y$ and $(s-c)=z$, then the following identity holds:

$$ab\cos{\gamma}=ab\cos^2{\frac{\gamma}{2}}-ab\sin^2{\frac{\gamma}{2}}=sz-xy$$

Substituting and multiplying by 4, 

$$4ab\cos{\gamma}=(a+b+c)(a+b-c)-(b+c-a)(a+c-b)$$

Simplifying,

$$2ab\cos{\gamma}=a^2+b^2-c^2$$

$\square$

A similar reasoning must show that $a^2=b^2+c^2-2bc\cos{\alpha}$ and $b^2=a^2+c^2-2ac\cos{\beta}$.

Acknowledgement. My sincerest thanks to Angina Seng for giving helpful comments which allowed me to simplify the proof.

Related material.

Several derivations of the Cosine Law, including Euclid's at cut-the-knot.