# 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.