Properties of Circles

Cyclic quadrilaterals: why opposite angles add to 180°

By Mr Gan · O-Level E Maths · Updated August 2026 · 7 min read

A cyclic quadrilateral is a four-sided shape with all four corners sitting on the same circle. The rule that solves nearly every exam question on it: opposite angles always add up to 180°. A second, related rule follows from it: an exterior angle equals the interior angle at the opposite vertex. Both rules come from the same circle theorem, so once you learn one, the other is nearly free.

Why students mix this up with other circle rules

By the time cyclic quadrilaterals show up, students have already met angle at the centre, angle in a semicircle, and angles in the same segment. It is easy to reach for the wrong tool: trying to find the centre of the circle when none is marked, or assuming all four angles are equal like in a rectangle. Neither approach is needed here.

The cyclic quadrilateral rule only cares about which pair of angles is opposite each other around the shape. Get the pairing right and the arithmetic is a single subtraction from 180°.

The method

Step 1

Confirm all 4 corners lie on the circle

Step 2

Identify the opposite angle pair

Step 3

Unknown = 180° − known opposite angle

Step 4

Check with the exterior angle rule

Watch the whole method in about a minute.

Worked example 1: finding an opposite angle

Quadrilateral PQRS has all four vertices on a circle. ∠P = 108° and ∠Q = 95°. Find ∠R and ∠S.

Solution

1
Going around the quadrilateral P, Q, R, S in order, the opposite pairs are (P, R) and (Q, S).
2
Since P and R are opposite: ∠R = 180° − ∠P = 180° − 108° = 72°.
3
Since Q and S are opposite: ∠S = 180° − ∠Q = 180° − 95° = 85°.
4
Check: all four angles should sum to 360° (true for any quadrilateral). 108 + 95 + 72 + 85 = 360°. ✓

Quick check: the four angles of any quadrilateral always add to 360°, cyclic or not. Use this as a fast sanity check on your two cyclic-quadrilateral answers before moving on.

Worked example 2: the exterior angle rule

In cyclic quadrilateral WXYZ, side WX is extended to a point T, forming exterior angle ∠TXY = 76°. Find ∠WZY, the interior angle opposite to X.

Solution

1
The exterior angle at a vertex of a cyclic quadrilateral equals the interior opposite angle. Here the exterior angle is at X, so the interior opposite angle is at Z.
2
Therefore ∠WZY = ∠TXY = 76°.
3
This matches the opposite-angle rule too: ∠WXY = 180° − 76° = 104°, and ∠WZY = 180° − ∠WXY = 180° − 104° = 76°. Same answer either way.

The step students get wrong

Students often use the exterior angle at a vertex against the interior angle at that same vertex instead of the opposite one. The exterior angle equals the interior angle at the vertex diagonally across the quadrilateral, not the one next to it. Sketch the shape and mark the opposite vertex before writing the equation.

Where this rule comes from

The cyclic quadrilateral rule is not a separate fact to memorise in isolation. It follows from the angle in the same segment theorem, which is covered in full in the circle theorems guide. Here is the short version. Draw both diagonals of cyclic quadrilateral ABCD. Diagonal AC splits ∠A into ∠DAC and ∠CAB. Angles in the same segment are equal, so ∠DAC = ∠DBC (both stand on chord DC) and ∠CAB = ∠CDB (both stand on chord BC). That makes ∠A = ∠DBC + ∠CDB, which are two of the three angles of triangle BCD. The third angle of that triangle is ∠BCD, which is just ∠C. So ∠A + ∠C is exactly the angle sum of a triangle, 180°.

Exam questions frequently combine cyclic quadrilaterals with triangle angle facts or isosceles triangle properties within the same circle diagram, so it helps to have your triangle congruence and angle reasoning solid too; see the congruence tests guide if that reasoning still feels shaky.

Worked example 3: combined with a triangle

In cyclic quadrilateral ABCD, diagonal AC is drawn. ∠ABC = 100°. In triangle ACD, ∠CAD = 35° and ∠ACD = 65°. Find ∠ADC two different ways.

1
Method A, inside the triangle: the three angles of triangle ACD sum to 180°, so ∠ADC = 180° − ∠CAD − ∠ACD = 180° − 35° − 65° = 80°.
2
Method B, cyclic quadrilateral rule: D is opposite B in quadrilateral ABCD, so ∠ADC = 180° − ∠ABC = 180° − 100° = 80°.
3
Both methods agree on ∠ADC = 80°. In an exam, whichever facts the question gives you determine which method is faster, but the two must always agree if the diagram is drawn correctly.

Exam tip: when a question gives you both a triangle inside the circle and cyclic quadrilateral information, work out the angle two different ways if you have time. Agreement between the two methods is strong evidence your answer is correct.


Frequently asked questions

Does the cyclic quadrilateral rule work for any four-sided shape?

No. It only applies when all four vertices lie on the same circle. A general quadrilateral's angles still sum to 360°, but opposite angles do not have to add to 180° unless it is cyclic. Always check the question states or the diagram shows all four points on the circle before applying this rule.

How do I know which two angles are "opposite" in the diagram?

Label the four vertices in order as you go around the shape, for example P, Q, R, S. The opposite pairs are the ones separated by one vertex on each side: P with R, and Q with S. They are never adjacent (next to each other) angles.

Is the exterior angle rule a separate theorem I need to memorise?

Not really; it is the same rule stated differently. Since the interior angle at a vertex and its exterior angle (on a straight line) add to 180°, and the interior angle also adds to 180° with its opposite interior angle, the exterior angle must equal that opposite interior angle. Learn the opposite-angles version well and this one follows automatically.

Can a square or rectangle be a cyclic quadrilateral?

Yes. Every rectangle (and therefore every square) is cyclic, because opposite angles are each 90°, and 90° + 90° = 180°. The four vertices of any rectangle always lie on a circle whose diameter is the rectangle's diagonal.

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