A tangent and a chord
Draw the tangent to a circle at a point P, and a chord from P to another point Q on the circle. The chord and the tangent meet at P and make two angles there, one on each side of the chord. The two angles make a straight line, so they add to 180°.
The chord PQ also cuts the circle into two segments. Pick one of the two angles between the tangent and the chord. It opens into one segment, the one on its own side of the chord. The segment on the other side of the chord is called the alternate segment for that angle.
The tangent at P and the chord PQ make an angle of 50° on the right of the chord, between the chord and the right-hand part of the tangent. The alternate segment for that angle is the shaded one, on the left. The angle at R, in the alternate segment, is 50° too.
The theorem
The angle between a tangent and a chord, at the point of contact, is equal to any angle at the circumference that stands on the chord from the alternate segment.
In the figure below, the angle between the tangent and the chord PQ is marked a. R is a point on the circle in the alternate segment, joined to P and to Q, and the angle PRQ is marked a as well.
The angle between the tangent at P and the chord PQ, and the angle at R in the alternate segment, are both marked a: they are equal.
First, a diameter
Suppose the chord from P passes through the center, so that it is a diameter. The tangent meets the radius at P at a right angle, and the diameter runs along that radius, so the angle between the tangent and the chord is 90°. The angle at R stands on a diameter, so it is an angle in a semicircle, which is also 90°. The two angles are equal.
Why it is true for any chord
Call the angle between the tangent and the chord t, on the side away from the center O, and join O to P and to Q.
The radius OP meets the tangent at a right angle, 90°. The chord PQ splits that right angle into t and the angle OPQ, so OPQ = 90° − t.
OP and OQ are both radii, so triangle OPQ is isosceles and its base angles are equal: OQP = 90° − t as well. Its three angles add to 180°, so the angle at the center is POQ = 180° − 2(90° − t) = 180° − 180° + 2t = 2t.
The angle PRQ stands on the same arc PQ as the angle POQ at the center, so by the angle at the center theorem it is half of it: PRQ . That is the tangent-chord angle.
In the figure, t = 50°, so OPQ = OQP = 90° − 50° = 40°, POQ = 180° − 40° − 40° = 100°, and the angle at R is 100° ÷ 2 = 50°.
The tangent at P meets the chord PQ at the angle t, on the right. The radius OP is at a right angle to the tangent, so the marked angle OPQ is 90° − t, and so is the marked angle OQP of the isosceles triangle OPQ. That leaves 2t at the center, and the angle at R is half of that, t.
The other angle, and the other segment
The theorem holds for the other angle between the tangent and the chord too. If t = 55°, the angle on the other side of the chord is 180° − 55° = 125°, and its alternate segment is the one R is not in. Take any point S there. PRQS is a cyclic quadrilateral, so its opposite angles at R and S add to 180°: the angle PSQ is 180° − 55° = 125°, equal to the other tangent-chord angle.
The same theorem with the tangent at another point and a different chord. The angle between the tangent and the chord, and the angle at R in the alternate segment, are again equal.
The usual mistakes
Looking in the wrong segment. The equal angle is at a point on the other side of the chord from the tangent-chord angle, not on the same side.
Matching the angle with an end of the chord. The tangent-chord angle at P equals the angle at a third point R on the circle, not the angle at Q, the other end of the chord.
Taking 90° − a or 180° − a. The theorem is an equality: if the tangent-chord angle is 40°, the angle in the alternate segment is 40°. The 90° belongs to the proof, and 180° − a is the angle on the other side of the chord.
Doubling. 2a is the angle at the center. R is on the circle, so its angle is a.
Worked example: A Model Car That Leaves a Circular Track Along the Tangent
Question A model car runs round a circular track and leaves it at the point T, driving on in a straight line TX along the tangent to the track at T. Two marker cones stand on the track at A and at B, on the other side of the chord TA from X. The angle between the car's path TX and the chord TA is 52°, and angle ATB is 63°. (a) A camera at B is pointed along BT and must turn to face A. Through what angle does it turn, that is, what is angle TBA? (b) Find angle TAB.
1.TX is the tangent at T and TA is a chord from the point of contact. The angle between them is 52°.
TX is the tangent at T and TA is a chord from T; the angle between them is 52°. 2.B lies in the alternate segment, on the other side of the chord TA from X. By the alternate segment theorem, angle TBA is equal to the angle between the tangent and the chord. (a) The camera turns through TBA = 52°.
(a) B is in the alternate segment, so TBA = 52°: the camera turns through 52°. 3.In triangle TAB, the angles add to 180°: TAB = 180° − 63° − 52°.
In triangle TAB the angles add to 180°: TAB = 180° − 63° − 52°. 4.(b) TAB = 65°. Check: continue the tangent the other way from T to a point Y. The angle between TY and the chord TB is 180° − 52° − 63° = 65°, and A lies in the alternate segment for that angle, so the theorem gives TAB = 65° by a second route.
(b) TAB = 65°. Check: the angle between TY and TB is 180° − 52° − 63° = 65°, and A is in its alternate segment.
Answer: (a) 52°; (b) 65°
Common mistakes
- Matching the 52° with the angle at A instead of the angle at B. The angle between the tangent and the chord TA equals the angle that TA makes at a point on the far side of TA, and that point is B; A is an end of the chord.
- Using 63° as the angle at B. Angle ATB is the angle at T between the two chords; the alternate segment theorem gives the angle at B from the tangent-chord angle of 52°, not from 63°.
More congruence, similarity and circle theorems problems, worked step by step →