# Directed Angles

**Directed angles** are angles that are directed. There are 2 ways of counting the directed angle $\measuredangle ABC:$

- the angle is positive when the points $A, B, C$ are in
*clockwise*order, and negative otherwise, or - the angle is positive when the lines $AB$ and $BC$ are in
*counterclockwise*order, and negative otherwise.

You should see that both ways are identical to each other. As an example, see the diagram below:

We take every angle modulo $180^\circ$, i.e.

$\cdots=-110^\circ=70^\circ=250^\circ=\cdots.$

This means that in the diagram above, $\measuredangle ABC = 50^\circ = 230 ^\circ = -130^\circ = \cdots$.

Directed angles might seem really annoying and useless at first, but soon you should find it natural and sometimes better than the normal angles.

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## Applications

Directed angles are useful in combining multiple cases in a statement into one. An example is shown below.

The theorem below has configuration issues; the two cases seem so different and hence we have to differentiate them. By using directed angles, we can significantly reduce the number of words used.

Theorem (Cyclic Quadrilaterals)Let $A, B, C, D$ be any four points, no three collinear.

(i) If $A$ and $B$ lie on the same side of $CD,$ then the four points are concyclic if and only if $∠CAD = ∠CBD.$

(ii) If $A$ and $B$ lie on different sides of $CD,$ then the four points are concyclic if and only if $∠CAD + ∠CBD = 180^\circ.$

Theorem (Directed Cyclic Quadrilaterals)Let $A, B, C, D$ be four points, no three collinear. Then they are concyclic if and only if

$\measuredangle CAD = \measuredangle CBD.$

*Task:* Make sure it works.

Let $X$ be any point. Points $A, B, C$ are collinear if and only if

$\measuredangle XBC = \measuredangle XBA.$

*Task:* Prove this yourself. $($Hint: Show that the assertion is equivalent to $\measuredangle ABC = 0.)$

Always remember that this doesn't work for **every single case**. You should always check your answer after writing it out. An example where directed angles will destroy the solution is shown below:

In the cyclic quadrilateral $ABCD$, let $I_1$ and $I_2$ denote the incenters of $\triangle{ABC}$ and ${DBC}$, respectively. Prove that $I_1I_2BC$ is cyclic.

Do not use directed angles; the problem is false if $A, C, B, D$ lie in that order.