What Is a Cyclic Quadrilateral?
A cyclic quadrilateral is a quadrilateral (a closed four-sided figure) whose four vertices all lie on the circumference of one circle. Because every vertex touches the circle, each of the four sides is a chord of that circle, and the circle is called the circumcircle of the quadrilateral. It is also known as an inscribed quadrilateral.
The one condition that matters is this: all four corners must touch the same circle. Draw any triangle and a circle always passes through its three vertices, so every triangle is "cyclic." A fourth vertex has no such guarantee, which is why some four-sided shapes are cyclic and others are not. A square, a rectangle, and an isosceles trapezium are cyclic; a general parallelogram and a non-square rhombus are not.
For Example : An astronomer once built the first trig tables using four points sitting on one circle.
Around 150 CE, Claudius Ptolemy needed a way to predict the positions of stars and planets. His breakthrough tool was a four-sided shape drawn inside a circle, and the relationship he found between its sides and diagonals still carries his name. That shape is the cyclic quadrilateral, and it turns a circle into a measuring instrument.
Why Do the Opposite Angles Add to 180°?
The defining property of a cyclic quadrilateral is that each pair of opposite angles is supplementary, meaning the two angles add to 180°. If the quadrilateral is $ABCD$, then
$$\angle A + \angle C = 180^\circ \qquad \text{and} \qquad \angle B + \angle D = 180^\circ.$$
Here is why it must be true. The proof rests on the inscribed angle theorem, which says an inscribed angle equals half the arc it opens onto.
$\angle A$ is inscribed in the circle and opens onto arc $BCD$, so $\angle A = \tfrac{1}{2}(\text{arc } BCD)$.
$\angle C$ opens onto the remaining arc $DAB$, so $\angle C = \tfrac{1}{2}(\text{arc } DAB)$.
Arc $BCD$ and arc $DAB$ together make the whole circle, which is $360^\circ$.
Adding the two angles:
$$\angle A + \angle C = \tfrac{1}{2}(\text{arc } BCD + \text{arc } DAB) = \tfrac{1}{2}(360^\circ) = 180^\circ.$$
The same argument works for $\angle B$ and $\angle D$. The converse is also true: if a quadrilateral's opposite angles add to 180°, its four vertices must lie on a common circle. That converse is the standard test for spotting a cyclic quadrilateral.
What Are the Properties of a Cyclic Quadrilateral?
Once the four vertices sit on a circle, several rules follow. These are the properties examiners test most.
Opposite angles are supplementary. $\angle A + \angle C = 180^\circ$ and $\angle B + \angle D = 180^\circ$.
Exterior angle equals the interior opposite angle. Extend one side; the exterior angle formed equals the interior angle at the opposite vertex.
Ptolemy's theorem. The product of the diagonals equals the sum of the products of the two pairs of opposite sides: $AC \times BD = (AB \times CD) + (BC \times AD)$.
Maximum area. Among all quadrilaterals with the same four side lengths, the cyclic one has the largest area (given by Brahmagupta's formula).
Perpendicular bisectors meet at the centre. The perpendicular bisector of every side passes through the circle's centre, since every side is a chord.
A quick reader question worth answering here:
Is a rectangle a cyclic quadrilateral? Yes. All four angles of a rectangle are 90°, so opposite angles sum to $90^\circ + 90^\circ = 180^\circ$, and a circle passes through all four corners with the centre at the diagonals' meeting point.
What Is the Area of a Cyclic Quadrilateral?
When only the four side lengths are known, the area of a cyclic quadrilateral comes from Brahmagupta's formula. Let the sides be $a$, $b$, $c$, $d$ and let $s$ be the semi-perimeter. Then
$$s = \frac{a + b + c + d}{2}, \qquad \text{Area} = \sqrt{(s - a)(s - b)(s - c)(s - d)}.$$
Variable key:
Symbol | Meaning |
|---|---|
$a, b, c, d$ | the four side lengths of the quadrilateral |
$s$ | the semi-perimeter, half the sum of the sides |
Area | the region enclosed, in square units |
This is the four-sided cousin of Heron's formula for a triangle. It only holds when the quadrilateral is cyclic, and it gives the maximum area any quadrilateral with those four side lengths can reach.
What Is the Circumradius of a Cyclic Quadrilateral?
Because all four vertices lie on one circle, that circle has a definite radius, called the circumradius $R$. Parameshvara's formula gives it directly from the four sides:
$$R = \frac{1}{4}\sqrt{\frac{(ab + cd)(ac + bd)(ad + bc)}{(s - a)(s - b)(s - c)(s - d)}}.$$
Here $a$, $b$, $c$, $d$ are the sides and $s$ is the same semi-perimeter used above. The denominator is exactly the quantity under Brahmagupta's square root, so the two formulas share the same building blocks: once the semi-perimeter is found, both the area and the circumradius follow.
Examples of Cyclic Quadrilaterals
The examples below move from a one-step angle check to a full diagonal calculation. Work each one before reading the solution.
Example 1
In cyclic quadrilateral $ABCD$, $\angle A = 95^\circ$. Find $\angle C$.
Opposite angles are supplementary.
$$\angle C = 180^\circ - \angle A = 180^\circ - 95^\circ = 85^\circ.$$
Final answer: $\angle C = 85^\circ$.
Example 2
In cyclic quadrilateral $PQRS$, $\angle P = 70^\circ$ and $\angle Q = 110^\circ$. A student says $\angle R = 110^\circ$ because "$Q$ and $R$ are next to each other." Is that right?
Watch the wrong path first. The student paired adjacent angles instead of opposite ones. Adjacent angles in a cyclic quadrilateral have no fixed sum, so that reasoning breaks.
The correct pairing uses opposite vertices. $R$ is opposite $P$, not $Q$.
$$\angle R = 180^\circ - \angle P = 180^\circ - 70^\circ = 110^\circ.$$
The number happens to match, but the reason is different, and relying on the wrong reason fails the moment the angles change.
Final answer: $\angle R = 110^\circ$, found from its opposite angle $\angle P$.
Example 3
One angle of a cyclic quadrilateral is 100°. Find the interior angle opposite to it and the exterior angle at that opposite vertex.
Interior opposite angle:
$$180^\circ - 100^\circ = 80^\circ.$$
The exterior angle at a vertex equals the interior angle at the opposite vertex, so the exterior angle here equals 100°.
Final answer: opposite interior angle $= 80^\circ$; exterior angle $= 100^\circ$.
Example 4
In cyclic quadrilateral $ABCD$, $\angle A = (2x + 4)^\circ$ and $\angle C = (3x + 6)^\circ$. Find $x$.
Set the opposite pair equal to 180°.
$$(2x + 4) + (3x + 6) = 180$$
$$5x + 10 = 180$$
$$5x = 170$$
$$x = 34.$$
Final answer: $x = 34$, giving $\angle A = 72^\circ$ and $\angle C = 108^\circ$.
Example 5
A cyclic quadrilateral has sides $AB = 6$, $BC = 8$, $CD = 6$, $AD = 8$ (a rectangle). Its diagonals are equal; find the product $AC \times BD$ using Ptolemy's theorem.
Ptolemy's theorem:
$$AC \times BD = (AB \times CD) + (BC \times AD)$$
$$AC \times BD = (6 \times 6) + (8 \times 8) = 36 + 64 = 100.$$
Since the diagonals of a rectangle are equal, $AC = BD$, so $AC^2 = 100$ and $AC = 10$. That matches the Pythagorean diagonal $\sqrt{6^2 + 8^2} = 10$.
Final answer: $AC \times BD = 100$, and each diagonal is 10.
Example 6
Prove that a cyclic parallelogram must be a rectangle.
In any parallelogram, opposite angles are equal, so $\angle A = \angle C$.
In a cyclic quadrilateral, opposite angles are supplementary, so $\angle A + \angle C = 180^\circ$.
Substitute $\angle A = \angle C$:
$$\angle A + \angle A = 180^\circ \implies 2\angle A = 180^\circ \implies \angle A = 90^\circ.$$
A parallelogram with a 90° angle is a rectangle.
Final answer: the only cyclic parallelogram is a rectangle.
Where Do Cyclic Quadrilaterals Show Up?
The idea earns its keep because it turns a hard measurement problem into an angle problem. Ptolemy built his chord tables, the ancestors of the sine table, by studying quadrilaterals inscribed in a circle, and his relationship let astronomers compute unknown distances across the sky from known ones. You can read the historical account of Ptolemy's theorem and its role in early trigonometry.
The same reasoning still runs through practical work today.
Surveying and navigation use the circle-through-known-points idea to fix an unknown position from measured angles.
Structural design relies on the maximum-area property when a fixed length of material must enclose the most space.
Circle theorems in exams lean on cyclic quadrilaterals to unlock angles that no other rule reaches.
The reason this shape matters is not that it looks neat. It is that a single condition, "all four corners on one circle," hands you a chain of exact angle and length relationships for free.
What Are the Most Common Mistakes With Cyclic Quadrilaterals?
Three errors cost the most marks. Each one comes from applying a real rule in the wrong place.
Mistake 1: Pairing adjacent angles instead of opposite ones
Where it slips in: angle-chase problems where two known angles happen to be next to each other.
Don't do this: assume the two neighbouring angles add to 180°. The most common first misstep is adding adjacent angles because they look like a pair.
The correct way: the supplementary rule applies only to opposite vertices. In $ABCD$, pair $A$ with $C$ and $B$ with $D$, never $A$ with $B$.
Mistake 2: Assuming every quadrilateral inscribed-looking shape is cyclic
Where it slips in: figures drawn freehand where a corner sits just off the circle.
Don't do this: apply the opposite-angle rule before checking that all four vertices genuinely touch the circle.
The correct way: confirm the shape is cyclic first, either from the given information or by the converse test (opposite angles summing to 180°). A general parallelogram and a non-square rhombus fail this test.
Mistake 3: Misreading Ptolemy's theorem
Where it slips in: length problems using the diagonal relationship.
Don't do this: write $AC \times BD = AB \times BC + CD \times AD$, mixing adjacent sides.
The correct way: pair opposite sides. It is $(AB \times CD) + (BC \times AD)$, the two pairs of opposite sides.
Conclusion
A cyclic quadrilateral has all four vertices on one circle, making every side a chord.
Opposite angles are always supplementary: $\angle A + \angle C = 180^\circ$.
The converse is the standard test: if opposite angles sum to 180°, the shape is cyclic.
Ptolemy's theorem links the diagonals to the two pairs of opposite sides.
The only cyclic parallelogram is a rectangle.
To work through cyclic-quadrilateral proofs with a teacher, explore Bhanzu's geometry tutor or a high school math tutor, and browse the full range of math classes online.
A Practical Next Step
Practice these problems to solidify your understanding: take any rectangle, confirm its opposite angles sum to 180°, then draw its circumcircle and check that the centre lands where the diagonals cross. Next, prove that an isosceles trapezium is cyclic using the converse test. If you get stuck on the angle pairing, return to the opposite-angle proof above. Want a live Bhanzu trainer to walk you through circle theorems step by step? Book a free demo class.
Read More
Quadrilaterals in geometry — the parent family of four-sided shapes.
Types of quadrilaterals — how squares, rectangles, and trapeziums relate.
Angles of a quadrilateral — why the four angles always sum to 360°.
Circles in geometry — chords, arcs, and the parts a cyclic quadrilateral is built on.
Central angle in geometry — the angle-arc relationship behind the proof.
Is a square a rectangle? — the classification that decides which parallelograms are cyclic.
Was this article helpful?
Your feedback helps us write better content