Law of Cosines: Proof Without Words I am trying to prove the Law of Cosines using the following diagram taken from Thomas' Calculus 11th edition.

I have an answer, but I think there must be a simpler or better way to do it.  Here is my answer:
Construct a coordinate system such that $(0,0)$ is located at the bottom right corner of the pictured triangle.  Then the red line intersects the hypotenuse at $(-a,0)$ and a leg at $(-b\cos\theta,b\sin\theta)$.  Thus the squared distance $c$ from $(-a,0)$ to $(-b\cos\theta,b\sin\theta)$ is
\begin{align}
c^2&=(-b\cos\theta-(-a))^2 + (b\sin\theta)^2\\
&=a^2-2ab\cos\theta+b^2\cos^2\theta+b^2\sin^2\theta\\
&=a^2+b^2-2ab\cos\theta.
\end{align}
I feel like there has to be a simpler way, since my proof is basically ignoring the right triangle, the circle, etc.  If somebody can show me another proof, that would be great.  Thanks.
UPDATE:  It looks like I needed the Intersecting Chords Theorem from Geometry to write $(a+c)(a-c)=(2a\cos\theta-b)(b)$.
 A: The image was a little difficult for me to parse at first, so here's a refinement:

Now ... With $A$ the vertex opposite $a$ in the $a$-$b$-$c$ triangle, we can express the power of $A$ with respect to the circle as two chord-chord products to get
$$(2a\cos\theta-b)\cdot b = (a-c)\cdot(a+c)$$
and the result follows. $\square$
A: In the similar triangles, the ratio of corresponding sides are equal. We have $$
\begin{array}{l}
\dfrac{2 a \cos \theta-b}{a-c}=\dfrac{a+c}{b} \\
c^{2}=a^{2}+b^{2}-2 a b\ cos \theta
\end{array}
$$

A: Let's consider triangle as vectors $ \overrightarrow{AC}=\overrightarrow{AB}+\overrightarrow{BC}$. Then we have $\overrightarrow{BC} = \overrightarrow{AC} -\overrightarrow{AB}$ and at last
$$\overrightarrow{BC^{2}} = \overrightarrow{AC^{2}} +\overrightarrow{AB^{2}}-2\cdot\overrightarrow{AC} \cdot \overrightarrow{AB} $$
Obtained equality written for length is Law of Cosines $c^2=a^2+b^2-2ab\cos\theta$. It can proved, that condition $\frac{a^2+b^2-c^2}{2ab} \in (-1,1)$ is necessary and sufficient for length $a,b,c$ to create triangle.
In Euclidean space there holds property, which can be considered as general form of cosines law
$$||x +y|| = ||x||^2+||y||^2+2(x,y)$$
where $(x,y)$ is scalar product.
