Alternate Segment Theorem

GCSE Circle Theorems circle theorems tangent chord
Practice this formula All formulas
\( \angle\text{ between tangent and chord }=\angle\text{ in opposite arc} \)

Statement (what it says)

If a tangent touches a circle at point \(A\) and \(AB\) is a chord through \(A\), then the angle between the tangent and the chord equals the angle in the opposite (alternate) segment of the circle:

\(\displaystyle \angle TAB = \angle ACB,\)

where the tangent line is \(TA\), chord is \(AB\), and \(C\) is any point of the circle on the arc of \(AB\) opposite the tangent (i.e. not containing \(A\)). In words: the angle you make by “skimming” the circle at \(A\) and turning onto chord \(AB\) equals the angle the chord \(AB\) subtends anywhere in the other segment of the circle.

Why it’s true (short reason)

  • The radius \(OA\) is perpendicular to the tangent at \(A\): \(OA \perp \text{tangent}\).
  • The angle at the centre is twice the angle at the circumference standing on the same arc: \(\angle AOB = 2\,\angle ACB\).
  • The “tangent–chord” angle \(\angle TAB\) equals the angle made by \(AB\) with a radius perpendicular to the tangent, which is half the central angle \(\angle AOB\). Hence \(\angle TAB = \angle ACB\).

What to look for in diagrams

You will usually see a tangent touching a circle at a point \(A\), and a chord \(AB\) starting at that same point. The target angle often has a little square/arc mark on the tangent–chord side or is labelled \(x^\circ\). The equal angle must be taken in the opposite segment of the circle (on the other side of chord \(AB\) from the tangent).

Recipe (how to use it)

  1. Mark the tangent point \(A\) and the chord \(AB\) through it.
  2. Identify the angle between the tangent and the chord at \(A\) (or the angle at the circumference that subtends \(AB\)).
  3. Move to the alternate segment: any angle on the circumference that subtends chord \(AB\) on the other side is equal.
  4. Finish the problem with usual tools: triangle angles sum to \(180^\circ\), isosceles/isosceles triangle facts, cyclic quadrilateral opposite angles \(180^\circ\), “angles in the same segment are equal”, etc.

Common pairings

  • Angles in the same segment are equal: If two angles subtend the same chord (on the same arc), they are equal.
  • Radius ⟂ tangent: Useful to build right angles around the touch point.
  • Cyclic quadrilateral: Opposite angles sum to \(180^\circ\).
  • Isosceles triangles from equal radii: If \(OA\) and \(OB\) are radii, then \(\triangle AOB\) is isosceles, giving base angles equal.

Mini proof sketch using angle–at–centre

Let \(O\) be the centre. Draw radius \(OA\) to the point of tangency and chord \(AB\). Because \(OA\perp\) tangent, the angle between the tangent and \(AB\) at \(A\) equals the angle between the radius through \(A\) and \(AB\) (they are complementary to the same angle). Meanwhile the angle at the circumference \( \angle ACB \) is half the central angle \( \angle AOB\). One can show (by equal right angles) that the tangent–chord angle is also half of \( \angle AOB\). Therefore \( \angle TAB = \angle ACB\).

Worked quick examples

  1. Given: Tangent at \(A\), chord \(AB\). The angle between tangent and \(AB\) is \(38^\circ\). Find: angle in the alternate segment. Answer: \(38^\circ\).
  2. Given: In \(\triangle ABC\) on the circle, tangent at \(A\) meets \(AB\). If \( \angle TAB = 32^\circ\). Find: \( \angle ACB \). Answer: \(32^\circ\).
  3. Given: \(\triangle ABC\) is inscribed; tangent at \(A\). If \( \angle ABC=68^\circ\) and \( \angle BAC=44^\circ\), then \( \angle ACB=68^\circ\) and equals the tangent–chord angle at \(A\).

Pitfalls

  • Picking the wrong angle at the circumference. The equal angle must sit on the arc opposite the tangent, not on the small segment next to \(A\).
  • Forgetting right angles. The radius to the tangency point is perpendicular to the tangent—often creates right triangles that unlock missing angles.
  • Mixing theorems. “Angle between tangent and chord equals angle in the alternate segment” is different from “angles in the same segment are equal”—though both may apply in the same diagram.

Exam strategy

  1. Mark the chord that starts at the tangency point.
  2. Copy the tangent–chord angle across to an angle on the other side of that chord (on the circumference).
  3. Complete the rest by triangle sums, isosceles, or cyclic-quadrilateral facts.
  4. Annotate each step with the theorem used (AST / same segment / radius ⟂ tangent / cyclic quad).

Summary

The alternate segment theorem links the “touch-and-turn” angle at a tangent to an equal angle on the opposite arc of the same chord. It is one of the quickest angle transfer tools in circle geometry—spot the tangent at \(A\), find chord \(AB\), and mirror the angle across to the far side.