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I am trying to prove the Law of Cosines using the following diagram taken from Thomas' Calculus 11th edition.

enter image description here

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 $(a+c)(a-c)=(2a\cos\theta-b)(b)$.

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  • $\begingroup$ Using the formula $a\cdot b=|a|\cdot |b|\cdot \cos(\gamma)$ , where $\gamma$ is the angle between the vectors $a$ and $b$, we can establish a "one-line proof", if you prefer this approach. $\endgroup$ – Peter Jul 13 at 10:22
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    $\begingroup$ Let $A$ be the vertex opposite $a$ in the $a$-$b$-$c$ triangle. Then express the power of $A$ with respect to the circle as two chord-chord products to get $$(2a\cos\theta-b)\cdot b = (a-c)\cdot(a+c)$$ $\endgroup$ – Blue Jul 13 at 10:28
  • $\begingroup$ @Blue That sounds like it ought to be an answer rather than a comment. $\endgroup$ – Arthur Jul 13 at 10:34
  • $\begingroup$ You might find this a bit of fun: researchgate.net/publication/… $\endgroup$ – Colin McDonagh Jul 13 at 10:57
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    $\begingroup$ @ColinMcDonagh: That's a good one. It's not immediately obvious that the trapezoidal area pairs are equal, but it works. ... Despite searching, I've never found a dissection proof that didn't have such compromises and/or was robust across a wide range of triangles. Ultimately, I decide that this trigonograph is about the best I could do. It's not a rigid dissection proof, but I like to think it has a certain charm. $\endgroup$ – Blue Jul 13 at 11:47
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The image was a little difficult for me to parse at first, so here's a refinement:

enter image description here

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. $\square$

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    $\begingroup$ Excellent. Impossible not to vote for your answer. $\endgroup$ – Sebastiano Jul 13 at 12:12
  • $\begingroup$ Note that the right triangle is somewhat necessary in order to find the length $2a\cos\theta$ of the red chord. $\endgroup$ – Arthur Jul 13 at 23:30
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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.

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