Cyclic quadrilaterals · Circle geometry

A cyclic quadrilateral is a four-sided shape with all four corners on a circle. Any quadrilateral has angles adding up to $360^\circ$ . A cyclic one has something extra.

See it for yourself

Cyclic quadrilateralDrag the corners

91 + 89 = 180°∠A + ∠C104 + 76 = 180°∠B + ∠D

Opposite angles are in the same colour. However you move the corners, ∠A + ∠C = 180° and ∠B + ∠D = 180°. This works only when all four corners are on the circle: that is what cyclic means.

Drag the corners around. Watch the two sums in the readout. They never change.
Make one angle very big. Its opposite angle shrinks to match.
Press Exterior angle at C and compare the gold angle outside C with $\angle A$ .

Opposite angles∠A + ∠C = 180°

Exterior angle∠DCE = ∠A

Why it’s true

This is lesson 1 again, used twice. Press Show why it works on the board.

$\angle A$ stands on arc BCD. By the angle at the centre, the angle at O on that arc is $2a$ .
$\angle C$ stands on the other arc, DAB. The angle at O on that arc is $2c$ .
Together those two centre angles go all the way round O, so $2a + 2c = 360^\circ$ .
Halve it: $a + c = 180^\circ$ .

The exterior angle follows at once. $\angle DCE$ and $\angle BCD$ are on a straight line, so $\angle DCE = 180^\circ - \angle BCD$ . And $\angle A = 180^\circ - \angle BCD$ too. So $\angle DCE = \angle A$ .

Be sure it really is cyclic

The rule only works when all four corners are on the circle. Two common traps:

A quadrilateral with one corner at the centre O is not cyclic. O is inside the circle, not on it.
Four points drawn near a circle aren’t enough. The question must say, or the diagram must show, that they lie on it.

Algebra with cyclic quadrilaterals

Many questions give angles as expressions. Write the fact as an equation, then solve it.

A past question, step by step

Worked example · WAEC 2024

WAEC 2024 · Paper 1 · Q30

In the diagram, $\angle SQR = 52^\circ$ and $\angle PRT = 16^\circ$ . Find the value of the angle marked $y$ .

The paper marks this diagram “not drawn to scale”; the redraw uses the true measurements.

Find the cyclic quadrilateral
P, Q, S and T lie on the circle, so PQST is a cyclic quadrilateral. The given $52^\circ$ is at Q, but outside the shape: it is between QS and QR, where QR carries the side PQ on past Q.
Think first. PQR is a straight line, so $\angle SQR$ is outside PQST. Which interior angle of PQST is opposite the corner Q?
Exterior angle = interior opposite angle
$\angle PTS = \angle SQR = 52^\circ \quad \text{(exterior angle of a cyclic quadrilateral)}$
Think first. You now know an angle at T in triangle PRT. What else do you know in that triangle?
Finish in triangle PRT
S lies on TR, so $\angle PTS$ is also the angle at T in triangle PRT. The angles of the triangle add up to $180^\circ$ :
$y = 180^\circ - 52^\circ - 16^\circ = 112^\circ$
The answer is C.

Your turn

JAMB 1990 · UME · Q37

Circle geometry

In the figure, $PQRS$ is a circle and $PQT$ and $SRT$ are straight lines. If $\angle SPQ = 81^\circ$ and $\angle PTS = 22^\circ$ , find $x = \angle PQR$ .

Worked solution (try it first)

The angles of triangle $PST$ add up to $180^\circ$ : $\angle PSR = 180^\circ - 81^\circ - 22^\circ$
$= 77^\circ$ .
$PQRS$ is a cyclic quadrilateral, and $\angle PQR$ is opposite $\angle PSR$ .
Opposite angles add up to $180^\circ$ : $x = 180^\circ - 77^\circ = 103^\circ$ , option C.

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