338. The Alternate Segment Secret · tangent-chord angle always equals alternate angle
∠BAQ = ∠ACB and ∠BAP = ∠ADB — each tangent-chord angle equals the inscribed angle in its alternate segment.
What this lesson covers
Try to break it
Drag A and B around the circle. The angle between the tangent at A and chord AB (∠BAQ) always equals the angle in the alternate segment ∠ACB. Try to break the equality; impossible. Edge case: if AB passes through O the chord becomes a diameter, ∠ACB = 90°, and the tangent is perpendicular to AB — 90° = 90° still holds.
How you build it
Draw a chord and a tangent at A.
- Mark the centre O of the circle.
- With centre O, draw a circle of radius about 220.
- Mark point A on the circle — this is the point of contact.
- Mark point B on the circle, away from A.
- Join A and B to form the chord AB.
- Draw the radius OA, joining the centre O to A.
- Construct the tangent PQ at A, perpendicular to radius OA.
- Mark point C on the major arc — the far side of chord AB from the tangent. The angle ∠ACB here equals the tangent-chord angle at A.
The proof, step by step
Prove that the angle between a tangent and a chord equals the angle in the alternate segment.
- In ΔABR, ∠ABR = 90° (Angle in a semicircle)
- ∠OAQ = 90° (Radius is perpendicular to tangent at point of contact)
- ∠RAB + ∠BAQ = 90° and ∠ARB + ∠RAB = 90°
- ∴ ∠ARB = ∠BAQ
- But ∠ARB = ∠ACB (Angles in the same segment)
- ∴ ∠BAQ = ∠ACB
Worked example
In the given figure, PQ is a tangent to a circle with centre O at point A. AB is a chord such that ∠BAQ = 65°. If C is a point on the major arc AB, find the measure of ∠ACB.
By the Alternate Segment Theorem, the angle between the tangent and chord (∠BAQ) equals the angle in the alternate segment (∠ACB). Thus, ∠ACB = ∠BAQ = 65°.
- 55°
- 65° — correct
- 75°
- 115°