Prove that $\angle A B C+\angle B C A+\angle C A B=0$ using directed angles

Question

I am reading EGMO and there author defines what a directed angle is and without giving any proof he listed all important properties of directed angle ,

I am seriously not getting comfortable with directed angles and how to use them . one of the properties that is listed is this

$$\angle A B C+\angle B C A+\angle C A B=0$$

Can someone give a proof of this fact?

Answer 1

You can think a directed angle as usual angle, except it encodes clockwise/anticlockwise in the sign and we always mod out 180 degrees.

Properly:

Definition 1: The directed angle $\angle(\ell,m)$ between two lines $\ell,m$ is the angle you rotate anticlockwise starting from $\ell$ and ending at $m$, modulo 180 degrees.

Definition 2: The directed angle $\angle ABC$ is defined as $\angle(\overline{AB},\overline{BC})$, where $\overline{AB}$ is the line $AB$, and similarly $\overline{BC}$.

So looking at your LHS: $\angle ABC$ is the (anticlockwise) angle turned to get from $\overline{AB}$ to $\overline{BC}$, $\angle BCA$ from $\overline{BC}$ to $\overline{CA}$, and finally $\angle CAB$ from $\overline{CA}$ to $\overline{AB}$. So we start from line $AB$ and get back to line $AB$, so must be a multiple of 180 degrees, which we mod out. So we get $0$ as in RHS.

Answer 2

In EGMO, directed angles are defined as angles $\mod {180}^{\circ}$, where $\measuredangle ABC$ to be positive if the vertices $A$, $B$, $C$ appear in clockwise order, and negative otherwise .

Note that in $\measuredangle ABC$, $\measuredangle BCA$, $\measuredangle CAB$ , the vertices are all either written in clockwise order or anticlockwise order .[why ?]

If they are all written in clockwise in clockwise order, then we have, $\measuredangle ABC+ \measuredangle BCA +\measuredangle CAB \equiv \angle A B C+\angle B C A+\angle C A B \equiv 180 \equiv 0 \mod 180^{\circ}$ [ by using the triangle sum property ]

If they are all written in anti-clockwise order, then we have, $\measuredangle ABC+ \measuredangle BCA +\measuredangle CAB \equiv -\angle A B C-\angle B C A-\angle C A B \equiv -(\angle A B C+\angle B C A+\angle C A B)\equiv -180^{\circ} \equiv 0 \mod 180^{\circ}$ [ by using the triangle sum property ]

And hence, we have $\measuredangle ABC+ \measuredangle BCA +\measuredangle CAB \equiv 0 \mod 180^{\circ} \implies \measuredangle ABC+ \measuredangle BCA +\measuredangle CAB = 0 $.