Arc and chord properties

Arcs and central angles. An arc is a piece of the circle's circumference cut off by two points; the minor arc is the shorter piece, the major arc the longer one. Every arc "subtends" an angle at the centre $O$ — the angle formed by joining its two endpoints to $O$. Larger central angles always correspond to longer arcs, so it is natural that equal central angles go with equal arcs:

where: $A, B, C, D$ are points on the circle, $O$ is the centre, and "arc $AB$" means the corresponding minor arcs (or both major arcs) measured on matching sides.

Chords and arcs. A chord is the straight line segment joining any two points on the circle (a diameter is the special case through the centre). Chords, too, are tied to arcs: equal chords cut off equal arcs, and conversely equal arcs are cut off by equal chords. This ICSE syllabus gives this result without proof:

where: $AB$ and $CD$ are chords of the same circle, or of circles with equal radii.

Why the three measures move together. Join a chord's endpoints to the centre and you get a triangle with two sides equal to the radius — for chord $AB$, triangle $OAB$ has $OA = OB = r$. A second chord $CD$ of the same length gives a second triangle $OCD$ with $OC = OD = r$ too. Because all three sides match ($OA = OC$, $OB = OD$, $AB = CD$), the triangles are congruent by SSS, so the angles they contain at $O$ must be equal: $\angle AOB = \angle COD$. That congruence is the hinge connecting chord length to central angle, and through the central-angle rule above, to arc length as well. The figure below shows the same central angle carved out twice by two equal chords, producing two equal arcs.

Equal chords force equal central angles, and this fact is the engine behind every arc-chord conversion in this topic, so it deserves to be pinned down as a full argument rather than taken on faith: given two chords of the same circle that are equal in length, we can show — using nothing beyond triangle congruence — that the angles they subtend at the centre must also be equal, which then hands us equal arcs for free through the central-angle-arc rule stated above.

Step 1. Set up the two triangles. Let $AB$ and $CD$ be chords of a circle with centre $O$ and radius $r$, with $AB = CD$. Join $OA$, $OB$, $OC$, $OD$. Since all four segments are radii of the same circle, $OA = OB = OC = OD = r$.

Step 2. Compare triangles $OAB$ and $OCD$. In these two triangles, $OA = OC$ (both radii), $OB = OD$ (both radii), and $AB = CD$ (given). All three pairs of corresponding sides are equal.

Step 3. Apply SSS congruence. Three matching pairs of sides means $\triangle OAB \cong \triangle OCD$ by the SSS (side-side-side) congruence rule. Congruent triangles have all corresponding angles equal, so the angle included between the two radii in each triangle must match: $\angle AOB = \angle COD$.

Step 4. Read off the arc conclusion. Since $\angle AOB = \angle COD$, the central-angle-arc rule gives arc $AB =$ arc $CD$ immediately — so equal chords produce equal central angles, which in turn produce equal arcs, tying all three quantities together in one chain of reasoning.

In a circle with centre $O$ and radius $7\text{ cm}$, chord $AB = 7\text{ cm}$ and chord $CD = 7\text{ cm}$, with $\angle AOB = 60^\circ$. Find $\angle COD$, state the relationship between arc $AB$ and arc $CD$, and find chord $EF$ of a second circle of the same radius if arc $EF$ subtends $60^\circ$ at its centre too.

Step 1. Use chord equality to get angle equality. Since $AB = CD = 7\text{ cm}$ are equal chords of the same circle, by the SSS argument above, $\triangle OAB \cong \triangle OCD$, so $\angle COD = \angle AOB = 60^\circ$.

Step 2. Use angle equality to get arc equality. Since $\angle AOB = \angle COD = 60^\circ$, the central-angle-arc rule gives arc $AB =$ arc $CD$.

Step 3. Cross-check with the equilateral triangle. Because $AB = OA = OB = 7\text{ cm}$, triangle $OAB$ has all three sides equal, making it equilateral, so all its angles are $60^\circ$ — confirming $\angle AOB = 60^\circ$ directly and matching the given data.

Step 4. Handle chord $EF$ in the second circle. Circle two has the same radius $7\text{ cm}$ as circle one, and arc $EF$ subtends $60^\circ$ at its centre — the same central angle as $\angle AOB$. Since the two circles are of equal radii and the central angles match, the chord-arc-angle equivalence applies across them too, so $EF = AB = \mathbf{7\text{ cm}}$.

Competitive stretch

Two arcs $PQ$ and $RS$ of the same circle subtend $80^\circ$ and $80^\circ$ at the centre, with chord $PQ = 9.2\text{ cm}$. By the same chain, $RS = PQ = 9.2\text{ cm}$. If instead arc $RS$ subtended $95^\circ \ne 80^\circ$, the central angles would differ, so the arcs would differ, so the chords could not be equal — a quick way to see how tightly the three quantities are bound together in both directions.

• Applying the equal-chords-equal-arcs rule across circles of different radii — it only holds within the same circle, or between circles of equal radius; a $7\text{ cm}$ chord in a small circle does not cut off the same arc as a $7\text{ cm}$ chord in a much bigger one.
• Comparing a minor arc on one side to a major arc on the other — "equal arcs" must compare corresponding arcs consistently, minor to minor or major to major.
• Pulling in the Class 10 result that the angle at the centre is twice the angle at the circumference — this topic only concerns angles measured at the centre, and inscribed-angle theorems are out of scope here.
• Remembering only the forward direction of a rule — both the equal-chords-give-equal-arcs statement and its converse (equal arcs give equal chords) hold, and problems often need the converse.

Ask the tutor: Want a walkthrough of the SSS congruence proof step by step, or a few more mixed chord-arc-angle practice problems?