viernes, 5 de junio de 2020

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


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:


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:
Substituting and multiplying by 4, 

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

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

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