Angle subtended by a chord

Draw any chord in a circle and join both its ends to the centre — you get an angle. Draw a second chord of the same length, join its ends to the centre too, and something clean happens: the two angles turn out exactly equal. This single fact, proved with nothing more than the SSS congruence rule already known from Class 9 triangles, opens the entire Circles chapter. It links a length (the chord) to an angle (at the centre) in both directions, so that from now on, "equal chords" and "equal central angles" can be swapped for each other freely in a proof. That swap is the real payoff: many circle problems become easy the moment you stop measuring lengths and start comparing the angles they subtend at the centre, or the other way round.

The core insight

Equal chords of a circle subtend equal angles at the centre, and conversely — so chord length and central angle can be traded for each other in any proof.

The core idea

Let a circle have centre $O$ . A chord $AB$ divides the circle's boundary into two arcs, and joining $O$ to $A$ and to $B$ creates the angle $\angle AOB$ — this is the angle subtended by the chord $AB$ at the centre. Every chord subtends exactly one such angle at the centre of its circle. The key result compares two chords of the same circle (or of two circles with equal radii, called congruent circles):

AB = CD \iff \angle AOB = \angle COD

where: $AB$ and $CD$ are chords of the same circle with centre $O$ , and $\angle AOB$ , $\angle COD$ are the angles they subtend at that centre. The double arrow means the statement works in both directions — equal chords force equal central angles, and equal central angles force equal chords, but only when both chords sit in the same circle or in circles of equal radius.

The forward direction (equal chords $\Rightarrow$ equal angles) is a theorem, proved below using triangle congruence.
The reverse direction (equal angles $\Rightarrow$ equal chords) is the converse, accepted and used without proof at this level.
Both directions break down the moment the two chords belong to circles of different radii — the theorem is strictly a same-circle (or congruent-circles) statement.

Figure

Two chords of equal length in a circle each joined to the centre showing two congruent central angles

Why it works

The equality is not a coincidence of drawing — it falls straight out of triangle congruence, so it is worth walking through the reasoning once carefully.

Derivation AEqual chords subtend equal angles at the centre

The goal is to show, using only the SSS (side-side-side) congruence criterion for triangles that a Class 9 student already knows, that if two chords of the same circle have equal length then the angles each one subtends at the centre must also be equal — turning a statement about lengths into a statement about angles with a single triangle-congruence argument.

Step 1. Set up the two triangles. Let $AB$ and $CD$ be chords of a circle with centre $O$ , and suppose $AB = CD$ . Join $OA$ , $OB$ , $OC$ , $OD$ . This creates two triangles, $\triangle AOB$ and $\triangle COD$ , both with one vertex at the centre.

Step 2. Match up three pairs of equal sides. $OA = OC$ , because both are radii of the same circle. $OB = OD$ , for the same reason — both are radii. And $AB = CD$ , which is exactly what was given. So all three sides of $\triangle AOB$ match all three sides of $\triangle COD$ .

Step 3. Apply SSS congruence. Since the three sides of $\triangle AOB$ equal the three corresponding sides of $\triangle COD$ (in the matching order $OA \leftrightarrow OC$ , $OB \leftrightarrow OD$ , $AB \leftrightarrow CD$ ), the SSS criterion gives $\triangle AOB \cong \triangle COD$ .

Step 4. Read off the equal angles. Corresponding parts of congruent triangles are equal (CPCT), and $\angle AOB$ and $\angle COD$ are the angles included between the radii in each triangle — they correspond to each other under the congruence. Hence $\angle AOB = \angle COD$ , which is exactly the claim: equal chords subtend equal angles at the centre. $\blacksquare$ The converse — if $\angle AOB = \angle COD$ then $AB = CD$ — is stated and used without a separate proof at this level; it is accepted as the "reverse read" of the same equivalence and is just as reliable for the same-circle case.

Worked example

Problem. $O$ is the centre of a circle. Chord $PQ = 6\text{ cm}$ subtends an angle of $70°$ at the centre, and chord $RS$ also subtends $70°$ at the centre of the same circle. Elsewhere in the same figure, chords $AB$ and $CD$ are equal, with $\angle AOB = 5x°$ and $\angle COD = (3x + 20)°$ . Find (a) the length of $RS$ , and (b) the value of $x$ and the measure of each of $\angle AOB$ and $\angle COD$ .

Step 1. Use the converse for $PQ$ and $RS$ . Both angles are subtended at the centre of the same circle: $\angle POQ = \angle ROS = 70°$ . By the converse, equal central angles in the same circle come from equal chords, so $PQ = RS$ .

Step 2. Read off the length. Since $PQ = 6\text{ cm}$ , this immediately gives $RS = 6\text{ cm}$ .

Step 3. Use the theorem for $AB$ and $CD$ , in the other direction. Here it is the chords that are given equal, $AB = CD$ , so by the theorem proved above, their central angles must be equal too: $\angle AOB = \angle COD$ , giving the equation $5x = 3x + 20$ .

Step 4. Solve for $x$ and the angles. $5x - 3x = 20 \implies 2x = 20 \implies x = 10$ . So $\angle AOB = 5(10) = 50°$ and $\angle COD = 3(10) + 20 = 50°$ — the two angles match, confirming the chords really are equal, as the theorem predicted.

Final answer: $RS = 6\text{ cm}$ , and $x = 10$ , with $\angle AOB = \angle COD = 50°$ .

Common traps

Mixing "same circle" with "different circles." The theorem and its converse both require the chords to lie in the same circle, or in circles of equal radii (congruent circles). Equal chords of circles with different radii do not need to subtend equal central angles.
Trying AA instead of SSS. Students sometimes reach for an angle-based congruence rule for $\triangle OAB$ and $\triangle OCD$ ; the correct route is SSS, using the two radii and the chord itself as the three matching sides.
Applying the converse across two separate circles. When a figure shows two different circles side by side, it is tempting to equate central angles to chords across them — this is only valid if the two circles are congruent (equal radii).
Confusing the centre angle with the circumference angle. $\angle AOB$ at the centre is a different angle from the one the same chord subtends at a point on the circle's remaining arc; that relationship (centre angle = twice the circumference angle) is a separate, later theorem.

Ask the tutor: if you want another worked example mixing the theorem with its converse, or a quick recap of SSS congruence before this proof, just ask the AI tutor about "angle subtended by a chord."

Does the theorem work if the two chords belong to two different circles?

Only if the two circles are congruent, meaning they have equal radii. If the radii differ, then even chords of the exact same length can subtend different-sized angles at their own centres, because a shorter radius makes the same chord "wrap around" more of the circle and subtend a larger angle. Always check "same circle or congruent circles" before applying either direction of the theorem.

Why do we need the converse separately if the theorem already links equal chords to equal angles?

The theorem only proves one direction: equal chords imply equal angles. In problems you are often given the angles are equal and asked to conclude the chords are equal — that is the reverse implication, the converse, and it needs to be invoked as its own accepted fact (it is not automatically true just because the forward direction is proved). Class 9 accepts it without demanding a separate SSS-style proof, but you should still name it explicitly when using it.

Is the angle subtended at the centre the same as the angle subtended on the circle itself?

No, they are different angles that happen to be related. The angle subtended at the centre, $\angle AOB$ , uses the centre $O$ as the vertex. The angle subtended at a point on the circle uses some other point on the remaining arc as the vertex, and that angle turns out to always be exactly half of the central angle — but that "double angle" result is a separate theorem you meet later in this chapter, not something to assume here.

Angle subtended by a chord

The core idea

Why it works

Worked example

Common traps

Before this

Where this leads