Suppose you have a triangle with vertices $a$, $b$, and $c$. I asked earlier how you can define the angles in a triangle based on the $\log$ function. I received the answer that, for instance, the angle at $a$ is found as $\left|\Im\log\left(\frac{c-a}{b-a}\right)\right|$.

Can this be used to show that the sum of angles in a triangle is $\pi$? I summed the angles as $$ \left|\Im\log\left(\frac{c-a}{b-a}\right)\right|+\left|\Im\log\left(\frac{a-b}{c-b}\right)\right|+\left|\Im\log\left(\frac{a-c}{b-c}\right)\right|. $$

I noticed that $\left|\Im\log\left(\frac{c-a}{b-a}\frac{a-b}{c-b}\frac{b-c}{a-c}\right)\right|=\left|\Im\log(-1)\right|=\pi$, when evaluating on the principal branch.

I had to cheat a bit and flip the $\frac{a-c}{b-c}$. Is there a more systematic way to prove this somehow?


Flipping the $\frac{a-c}{b-c}$ (or any of the fractions in your angle expressions) actually isn't cheating. First, perhaps more intuitively, when you said that the measure of angle $a$ is $\left|\Im\log\left(\frac{c-a}{b-a}\right)\right|$, swapping $b$ and $c$ should not change the measure of the angle, so you should expect $\left|\Im\log\left(\frac{c-a}{b-a}\right)\right|=\left|\Im\log\left(\frac{b-a}{c-a}\right)\right|$. More formally, $\frac{b-a}{c-a}=(\frac{c-a}{b-a})^{-1}$, so $$\begin{align} \left|\Im\log\left(\frac{b-a}{c-a}\right)\right|&=\left|\Im\log\left(\left(\frac{c-a}{b-a}\right)^{-1}\right)\right| \\ &=\left|-\Im\log\left(\frac{c-a}{b-a}\right)\right| \\ &=\left|\Im\log\left(\frac{c-a}{b-a}\right)\right| \end{align}$$ (since the factor of $-1$ only changes the sign of the imaginary part, and that sign change is wiped out by the absolute value).

What might be cheating, though, is combining the absolute values.

I'd go about this in a slightly different way. Let's start by backing up to $$\begin{align} m\angle a&=\left|\Im\log\left(\frac{c-a}{b-a}\right)\right| \\ &=\left|\Im\left(\log(c-a)-\log(b-a)\right)\right| \\ &=\left|\Im\log(c-a)-\Im\log(b-a)\right|. \end{align}$$ $\Im\log(c-a)$ and $\Im\log(b-a)$ are the directed angles from the positive real axis to the ray from $0$ to $c-a$ and $b-a$, respectively, so $\Im\log(c-a)-\Im\log(b-a)$ is the directed angle from $b-a$ to $c-a$. When I say "directed" angle, I mean that a positive angle is a counterclockwise rotation.

Now, without loss of generality, let the vertices be labeled $a$, $b$, and $c$ in a counterclockwise direction around the triangle:

diagram of triangle

Working carefully, we can ensure that we measure each angle in the positive direction, and thus avoid the absolute values: $$\begin{align} m\angle a&=\Im\log(c-a)-\Im\log(b-a)=\Im\log\frac{c-a}{b-a} \\ m\angle b&=\Im\log(a-b)-\Im\log(c-b)=\Im\log\frac{a-b}{c-b} \\ m\angle c&=\Im\log(b-c)-\Im\log(a-c)=\Im\log\frac{b-c}{a-c} \\ \\ m\angle a+m\angle b+m\angle c&=\Im\log\frac{c-a}{b-a}+\Im\log\frac{a-b}{c-b}+\Im\log\frac{b-c}{a-c} \\ &=\Im\log\left(\frac{c-a}{b-a}\cdot\frac{a-b}{c-b}\cdot\frac{b-c}{a-c}\right) \\ &=\Im\log\left((-1)^3\right) \\ &=\pi. \end{align}$$

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