Cyclic Quadrilaterals is the habit of using opposite angles and equal subtended arcs when four points lie on one circle. 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.
$$ABCD \text{ cyclic} \;\Longleftrightarrow\; \angle A + \angle C = \angle B + \angle D = 180^\circ$$
Four points lie on one circle exactly when each pair of opposite angles of the quadrilateral adds to a straight angle.
Why It Works
1
Inscribe $ABCD$ in a circle. The inscribed angle $\angle A$ subtends arc $BCD$ and $\angle C$ subtends arc $DAB$; together these arcs make the whole circle.
2
Each inscribed angle is half its subtended arc, so $\angle A + \angle C = \tfrac12(\text{arc }BCD) + \tfrac12(\text{arc }DAB) = \tfrac12(360^\circ) = 180^\circ$.
3
The same holds for $\angle B + \angle D$, so opposite angles are supplementary.
4
The converse is also true: if opposite angles sum to $180^\circ$, the four points are concyclic — a key tool for proving points lie on a circle.
Worked Examples
Example 1
Quadrilateral $ABCD$ is inscribed in a circle. If $\angle A = 95^\circ$ and $\angle B = 70^\circ$, find $\angle C$ and $\angle D$.
1
Opposite angles of a cyclic quadrilateral are supplementary.
2
$\angle C = 180^\circ - \angle A = 180^\circ - 95^\circ = 85^\circ$.
3
$\angle D = 180^\circ - \angle B = 180^\circ - 70^\circ = 110^\circ$.
Answer:
$\angle C = 85^\circ$, $\angle D = 110^\circ$
Warm-up
In cyclic quadrilateral $ABCD$, the diagonal $BD$ is drawn. If $\angle BAC = 40^\circ$ and $\angle ACB = 65^\circ$, find $\angle ADB$.
1
Inscribed angles subtending the same arc are equal. The angles $\angle ADB$ and $\angle ACB$ both subtend arc $AB$.
2
Therefore $\angle ADB = \angle ACB = 65^\circ$ directly (the $\angle BAC$ value is a distractor).
Answer:
$\angle ADB = 65^\circ$
Contest level
In $\triangle ABC$ the altitudes from $B$ and $C$ meet the opposite sides at $E$ and $F$. Show that $B$, $C$, $E$, $F$ lie on a circle, and find its diameter in terms of the triangle.
1
$\angle BEC = 90^\circ$ (altitude from $B$ is perpendicular to $AC$ at $E$) and $\angle BFC = 90^\circ$ (altitude from $C$ at $F$).
2
Points $E$ and $F$ both see segment $BC$ at a right angle, so each lies on the circle with diameter $BC$ (Thales' converse).
3
Hence $B, C, E, F$ are concyclic, on the circle whose diameter is $BC$.
Answer:
They are concyclic on the circle with diameter $BC$ (the 'feet of altitudes' circle).
Olympiad / Challenge
Two circles meet at $P$ and $Q$. A line through $P$ meets the first circle again at $A$ and the second at $B$; a line through $Q$ meets the first at $C$ and the second at $D$. Prove $AC \parallel BD$.
1
Work with the angles each line makes with the common chord $PQ$. On the first circle $A, P, Q, C$ are concyclic, so the inscribed angles $\angle CAP$ and $\angle CQP$ subtend the same arc $CP$, giving $\angle PAC = \angle PQC$.
2
On the second circle $B, P, Q, D$ are concyclic, so $\angle PBD = \angle PQD$ (inscribed angles on arc $PD$).
3
Since $C, Q, D$ are collinear, $\angle PQC$ and $\angle PQD$ are supplementary; combined with the previous two equalities, line $AC$ and line $BD$ make equal (alternate) angles with the transversal $APB$.
4
Equal alternate angles force $AC \parallel BD$.
Answer:
$AC \parallel BD$ (a classic two-circle parallel-lines lemma).
Going Deeper
Generalization: 'opposite angles supplementary' is one of FOUR equivalent concyclicity tests, all interchangeable: (1) opposite angles sum to $180^\circ$; (2) an exterior angle equals the opposite interior angle; (3) two points see a segment at equal angles on the same side; (4) Ptolemy's equality $AC\cdot BD = AB\cdot CD + AD\cdot BC$ holds. Pick whichever matches the data you have.
Where it appears: spotting a hidden cyclic quadrilateral is the single most common breakthrough in olympiad geometry — it instantly imports the whole inscribed-angle toolkit. Right angles are a giveaway (Thales: any right angle on a segment puts the vertex on the circle with that segment as diameter).
Pitfall: the supplementary-angle test requires the quadrilateral to be CONVEX with vertices in cyclic order $A, B, C, D$. If the points are not in order around the circle (a self-crossing 'quadrilateral'), opposite angles need not sum to $180^\circ$; check the configuration before applying the test.
Spot the Signal
Look for problems where the key step is using opposite angles and equal subtended arcs when four points lie on one circle.
You can describe the hard part as using opposite angles and equal subtended arcs when four points lie on one circle, 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 cyclic quadrilaterals, then rewrite the givens around it.
Name the relation that makes Cyclic Quadrilaterals 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 Cyclic Quadrilaterals just because the surface looks familiar; verify the required condition first.
Applying Cyclic Quadrilaterals 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 cyclic quadrilaterals reveal the geometric configuration; then compute only what remains.
Try it on:
Practice a contest problem where the key step is using opposite angles and equal subtended arcs when four points lie on one circle.