Cyclic quadrilaterals

Take any four points on a circle, in order, and join them up — you get a cyclic quadrilateral. That single condition, all four corners resting on one circle, forces a strict rule on the shape's angles: each pair of opposite angles must add to exactly $180°$ , no matter how the four points are chosen. This closes out the Class 9 circle-theorem sequence, and it works in both directions — the angle rule also gives a fast test for whether four points can be circled by one circle at all, without ever drawing that circle. It connects back to two earlier results: angles in the same segment are equal, and the angle at the centre is double the angle at the circumference. Board questions mostly ask you to solve for unknown angles using algebraic expressions, so speed with linear equations matters as much as knowing the rule.

The core insight

In a cyclic quadrilateral, opposite angles always sum to $180°$ , and — the converse — any quadrilateral whose opposite angles sum to $180°$ must be cyclic.

The core idea

A cyclic quadrilateral $ABCD$ is a quadrilateral whose four vertices $A$ , $B$ , $C$ , $D$ all lie on a single circle. The circle is called the circumscribing circle of the quadrilateral.

The main theorem. The sum of each pair of opposite angles is $180°$ :

\angle A + \angle C = 180°, \qquad \angle B + \angle D = 180°

where: $\angle A$ and $\angle C$ are the two vertex angles that sit diagonally across from one another (not sharing a side), and likewise $\angle B$ and $\angle D$ . This is one of three "state without proof" facts prescribed for this NCERT unit, alongside its converse and the fact that exactly one circle passes through any three non-collinear points.

The converse. If a quadrilateral has one pair of opposite angles summing to $180°$ , it is automatically cyclic — its four vertices lie on a common circle. This converse is what makes the theorem useful rather than just descriptive: given four points with no circle drawn, checking one angle sum tells you whether they are concyclic.

A companion test for four points. If a segment $AB$ subtends equal angles at two points $C$ and $D$ on the same side of $AB$ , i.e. $\angle ACB = \angle ADB$ , then $A$ , $B$ , $C$ , $D$ are concyclic. This is the natural partner of "angles in the same segment are equal," and it is often the first move in a proof, establishing that four points lie on a circle before the opposite-angle-sum rule is applied to that quadrilateral.

Why a fourth point either fits or it doesn't. Any three non-collinear points determine exactly one circle. So once $A$ , $B$ , $C$ are fixed, there is one specific circle through them, and a fourth point $D$ is either on it (the quadrilateral is cyclic) or it isn't — "concyclic" is a sharp, checkable condition, not a vague one.

Figure

A cyclic quadrilateral ABCD inscribed in a circle with opposite angles A and C marked and their sum shown as 180 degrees

Derivation AOpposite angles of a cyclic quadrilateral sum to 180 degrees

The goal is to show, using only the earlier theorem that the angle an arc subtends at the centre is twice the angle it subtends anywhere on the remaining circumference, that the two opposite angles of a cyclic quadrilateral inscribed in a circle must add up to exactly a straight angle of $180°$ , with no exceptions for how the four points are placed.

Step 1. Look at vertex $A$ of cyclic quadrilateral $ABCD$ . The angle $\angle A$ (that is, $\angle DAB$ ) is the angle inscribed at $A$ , standing on the arc $BCD$ — the arc that goes from $B$ to $D$ the long way round, passing through $C$ , and not containing $A$ itself.

Step 2. By the centre-circumference theorem, the angle this arc $BCD$ subtends at the centre of the circle is exactly twice $\angle A$ . Call the arc's central angle $2\angle A$ .

Step 3. Now look at vertex $C$ . By the same reasoning, $\angle C$ (that is, $\angle BCD$ ) is the angle inscribed at $C$ , standing on the arc $BAD$ — the arc from $B$ to $D$ the other way round, passing through $A$ . Its central angle is $2\angle C$ .

Step 4. The two arcs $BCD$ and $BAD$ together make up the entire circle exactly once, since every point of the circle lies on one arc or the other with $B$ and $D$ as the shared endpoints. So their central angles must add to a full turn: $2\angle A + 2\angle C = 360°$ .

Step 5. Divide both sides by $2$ : $\angle A + \angle C = 180°$ . The identical argument applied to vertices $B$ and $D$ , using arcs $ABC$ and $ADC$ , gives $\angle B + \angle D = 180°$ . Both opposite pairs are forced to be supplementary, and the result holds for every cyclic quadrilateral, not just special ones.

Worked example

Problem. $ABCD$ is a cyclic quadrilateral in which $\angle A = 3x - 5°$ and $\angle C = 2x + 15°$ . Find $x$ and the values of $\angle A$ and $\angle C$ .

Step 1. Since $ABCD$ is cyclic, $\angle A$ and $\angle C$ are opposite angles, so they must sum to $180°$ :

(3x - 5°) + (2x + 15°) = 180°

Step 2. Collect like terms. The $x$ -terms give $3x + 2x = 5x$ , and the constants give $-5° + 15° = 10°$ :

5x + 10° = 180°

Step 3. Solve for $x$ : subtract $10°$ from both sides to get $5x = 170°$ , then divide by $5$ to get $x = 34°$ .

Step 4. Substitute $x = 34°$ back into each expression: $\angle A = 3(34°) - 5° = 102° - 5° = 97°$ , and $\angle C = 2(34°) + 15° = 68° + 15° = 83°$ .

Step 5. Check the answer against the theorem: $97° + 83° = 180°$ , exactly as required. Answer: $x = 34°$ , $\angle A = 97°$ , $\angle C = 83°$ .

Competitive stretch

Now suppose the same quadrilateral $ABCD$ has $\angle B = 95°$ . Since $\angle B$ and $\angle D$ are the other opposite pair, $\angle D = 180° - 95° = 85°$ — you never need to compute it from scratch once one pair is known, because all four angles of any quadrilateral sum to $360°$ , and the first pair already accounts for $180°$ of that total. A quick full check: $97° + 95° + 83° + 85° = 360°$ , confirming everything is consistent.

Common traps

Mixing up opposite and adjacent angles. In $ABCD$ , the opposite pairs are $(\angle A, \angle C)$ and $(\angle B, \angle D)$ — not neighbouring angles like $(\angle A, \angle B)$ . Only opposite pairs are guaranteed to sum to $180°$ .
Assuming any quadrilateral qualifies. A general quadrilateral's four angles sum to $360°$ in total, but its opposite pairs need not each be $180°$ — that stronger property is special to cyclic quadrilaterals.
Applying the concyclic test with points on the wrong side. "Equal angles subtended at two points on the same side of the segment" is the precise condition for concyclic points; points on opposite sides instead give supplementary angles, which is the cyclic-quadrilateral case, not this one.
Solving both opposite pairs from scratch. Once one pair of opposite angles is confirmed to sum to $180°$ , the other pair is automatically forced to $180°$ too, since all four angles of any quadrilateral total $360°$ .

Ask the tutor: want to practice spotting cyclic quadrilaterals inside mixed circle-and-triangle diagrams, or trace how this theorem grows out of the centre-circumference angle rule?

Why do opposite angles have to sum to 180 degrees instead of some other number?

It traces back to the centre-circumference theorem. Each opposite angle sits on one of the two arcs cut off by the diagonal-forming vertices, and those two arcs together make the whole circle, so their central angles add to $360°$ . Since each inscribed angle is half its arc's central angle, the two inscribed angles add to half of $360°$ , which is $180°$ . It is not a coincidence — it is forced by the arc geometry every single time.

Do I need to prove the cyclic quadrilateral theorem in my exam?

No. NCERT Class 9 lists this as a "state and use" result, along with its converse and the fact that three non-collinear points determine one circle. You are expected to quote the theorem correctly and apply it to solve for angles, not to reproduce the arc-based reasoning — though understanding it, as shown above, makes the rule much easier to remember and to apply confidently in mixed diagrams.

How is the concyclic-points test different from the cyclic quadrilateral theorem itself?

The concyclic-points test looks at a single segment and two points that see it at equal angles from the same side — that tells you those four points sit on one circle. The cyclic quadrilateral theorem is what you use after you know the four points are already concyclic: it tells you their opposite angles sum to $180°$ . In many proofs you use the test first to establish concyclicity, then the theorem to extract an angle value.

Cyclic quadrilaterals

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads