Angle Chasing is the habit of tracking equal, complementary, supplementary, and cyclic angles until the target angle appears. In contest math, that habit turns a crowded setup into a relation the student can test, bound, count, or compute. MathGrit teaches it as a recognizable signal, a deliberate move, and a final translation back to the original question.
$$\text{Inscribed Angle Theorem: } \angle\text{(inscribed)} = \tfrac{1}{2}\,\angle\text{(central on the same arc)}$$
You track equal, supplementary, and inscribed angles step by step around a figure until the angle you want is forced.
Why It Works
1
Three facts drive almost every chase: angles in a triangle sum to $180^\circ$, angles on a straight line sum to $180^\circ$, and an inscribed angle is half the central angle subtending the same arc.
2
From the inscribed angle theorem, two inscribed angles subtending the SAME arc are equal — this lets you transport an angle across a circle.
3
A tangent and a chord make an angle equal to the inscribed angle in the alternate segment; parallel lines give equal alternate/corresponding angles.
4
Label one unknown angle $x$, then repeatedly apply these rules to express every other angle in terms of $x$ until a known total (like $180^\circ$) pins $x$ down.
Worked Examples
Example 1
In $\triangle ABC$, $AB = AC$ and $\angle BAC = 40^\circ$. Point $D$ lies on $BC$ so that $AD$ bisects $\angle BAC$. Find $\angle ADB$.
1
Since $\triangle ABC$ is isosceles with $AB = AC$, the base angles are equal: $\angle ABC = \angle ACB = \tfrac{180^\circ - 40^\circ}{2} = 70^\circ$.
2
The bisector $AD$ splits $\angle BAC$ into two $20^\circ$ angles, so $\angle BAD = 20^\circ$.
3
In $\triangle ABD$ the angles sum to $180^\circ$: $\angle ADB = 180^\circ - \angle ABD - \angle BAD = 180^\circ - 70^\circ - 20^\circ = 90^\circ$.
Answer:
$\angle ADB = 90^\circ$
Warm-up
Two chords $AB$ and $CD$ of a circle meet at an interior point $P$. The intercepted arcs satisfy $\text{arc } AC = 70^\circ$ and $\text{arc } BD = 30^\circ$. Find $\angle APC$.
1
An angle formed by two chords meeting inside a circle equals half the SUM of the two intercepted arcs (the arc it opens onto and the arc opposite).
2
Here $\angle APC$ opens onto arc $AC$ and its vertical angle opens onto arc $BD$, so $\angle APC = \tfrac12(\text{arc } AC + \text{arc } BD)$.
In $\triangle ABC$, $\angle B = 60^\circ$ and $\angle C = 70^\circ$. The altitude from $A$ meets $BC$ at $H$ and the angle bisector from $A$ meets $BC$ at $D$. Find $\angle HAD$.
1
First $\angle A = 180^\circ - 60^\circ - 70^\circ = 50^\circ$, so the bisector gives $\angle BAD = \angle DAC = 25^\circ$.
2
In right triangle $\triangle ABH$ (right angle at $H$), $\angle BAH = 90^\circ - \angle B = 90^\circ - 60^\circ = 30^\circ$.
3
Both $H$ and $D$ lie on $BC$ with $D$ on the larger-angle side, and $\angle HAD = |\angle BAD - \angle BAH| = |25^\circ - 30^\circ| = 5^\circ$.
Answer:
$\angle HAD = 5^\circ$ (in general $\tfrac12|\angle B - \angle C|$, here $\tfrac12|60^\circ - 70^\circ| = 5^\circ$).
Olympiad / Challenge
Let $I$ be the incenter of $\triangle ABC$. Prove that $\angle BIC = 90^\circ + \tfrac12\angle A$, and evaluate it when $\angle A = 40^\circ$.
1
Let $\angle A = \alpha$, $\angle B = \beta$, $\angle C = \gamma$, so $\alpha + \beta + \gamma = 180^\circ$.
2
The incenter lies on all three bisectors, so in $\triangle BIC$ the base angles are $\angle IBC = \tfrac{\beta}{2}$ and $\angle ICB = \tfrac{\gamma}{2}$.
$\angle BIC = 90^\circ + \tfrac12\angle A$; for $\angle A = 40^\circ$ this is $110^\circ$.
Going Deeper
Generalization: every chord-and-secant angle is half a sum or half a difference of arcs. Inscribed angle = $\tfrac12$(arc); two chords crossing inside = $\tfrac12$(sum of the two intercepted arcs); two secants/tangents meeting OUTSIDE = $\tfrac12$(difference of arcs). Memorizing this one family replaces a dozen separate rules.
Where it appears: angle chasing is the universal first move on any circle/triangle configuration problem (AMC 10/12 geometry, early AIME). It is also the standard way to PROVE concyclicity — if you can chase two angles to be equal that subtend the same segment, the four points lie on a circle.
Pitfall: an inscribed angle is half the central angle only for the SAME arc, and only when the vertex is on the major (correct) arc. If the vertex is on the other arc you get the supplement. Always confirm which side of the chord your vertex lies on before equating angles.
Spot the Signal
Look for problems where the key step is tracking equal, complementary, supplementary, and cyclic angles until the target angle appears.
You can describe the hard part as tracking equal, complementary, supplementary, and cyclic angles until the target angle appears, but a direct attack starts producing clutter.
The problem rewards preserving structure instead of expanding, listing, or guessing too early.
Learn the Move
Start by identify the diagram relation that calls for angle chasing, then rewrite the givens around it.
Name the relation that makes Angle Chasing legal before doing computation.
Use the new relation to replace the messiest part of the problem with a cleaner one.
Translate the result back to the quantity the problem actually asks for.
Avoid These Traps
Do not use Angle Chasing just because the surface looks familiar; verify the required condition first.
Applying Angle Chasing because it sounds relevant, without checking the trigger first.
Stopping after spotting the technique instead of finishing the calculation or proof.
MathGrit Coach Note
Let angle chasing reveal the geometric configuration; then compute only what remains.
Try it on:
Practice a contest problem where the key step is tracking equal, complementary, supplementary, and cyclic angles until the target angle appears.