Angles in the Same Segment

Statement

The circle theorem known as angles in the same segment states that angles subtended at the circumference of a circle by the same chord are equal. Put simply, if you pick two points on a circle and join them to form a chord, then any angle you make at the circumference using that chord will have the same size, provided the angles are in the same segment of the circle.

\[ \angle APB = \angle AQB \]

Here, \(AB\) is a chord, and \(P\) and \(Q\) are two points on the same arc opposite that chord.

Why it’s true (short reason)

• The proof involves drawing the radii from the circle’s centre and splitting the chord into isosceles triangles.
• It can also be explained using the fact that the angle at the centre is twice the angle at the circumference subtended by the same arc. Since both angles subtended by the chord share the same arc, they must be equal.

Recipe (how to use it)

1. Identify a chord in the circle (two points joined on the circle).
2. Mark two angles that subtend this chord at the circumference in the same segment.
3. State that the two angles are equal.
4. Use the equality to form equations and solve for unknowns.

Spotting it

• Look for a chord drawn across the circle with multiple points marked on the circumference in the same arc.
• Often shown as a cyclic quadrilateral with two opposite angles clearly formed by the same chord.
• Key clue: arrows or dots at the circumference showing angles standing on the same chord.

Common pairings

• Angle at the centre: often used alongside the theorem that the centre angle is twice the circumference angle.
• Cyclic quadrilaterals: opposite angles in a cyclic quadrilateral add to 180°, and these often appear together with equal angles in the same segment.
• Tangent-chord angle: the alternate segment theorem pairs with equal-in-segment facts in multi-step problems.

Mini examples

1. Given: chord \(AB\) subtends \(\angle ACB=38^{\circ}\). Find: \(\angle ADB\). Answer: \(38^{\circ}\).
2. Given: chord \(PQ\) subtends equal angles at \(R\) and \(S\). If \(\angle PRQ=72^{\circ}\), find: \(\angle PSQ\). Answer: \(72^{\circ}\).
3. Given: chord \(XY\) subtends equal angles at \(Z\) and \(W\). One angle is marked \(4x\), the other \(60^{\circ}\). Find: \(x\). Answer: \(x=15\).

Pitfalls

• Mixing arcs: the angles must be in the same segment (i.e., on the same side of the chord). Angles in opposite segments are not equal.
• Confusing with alternate segment theorem: tangent-chord problems use a different rule; do not apply the “same segment” theorem unless both angles are at the circumference.
• Forgetting the chord: both equal angles must be subtended by exactly the same chord, not just “similar-looking” arcs.
• Exterior points: ensure the angle is at the circumference, not outside the circle.

Exam strategy

• Mark equal angles with the same symbol immediately when you spot the chord and same-segment situation.
• If algebraic expressions are used (e.g., \(3x+10\) and \(50\)), set them equal and solve quickly.
• Remember to justify: always write “angles in the same segment are equal” when giving reasons.
• Check you are on the correct side of the chord: all equal angles must lie on the same arc.

Summary

Angles subtended at the circumference of a circle by the same chord are equal, provided they are in the same segment. This theorem is a quick way to link two or more angles in circle geometry, and it regularly appears in GCSE exam questions, often in combination with other circle theorems. Spot the chord, check the arc, then state the angles are equal and proceed to solve.