The alternate segment theorem — Mathematics

Before this

This leads to

Combining circle theorems
Similar triangles · coming soon
Tangents and normals to a circle · coming soon

An angle made by a line that never goes near it

Draw a circle and a tangent that touches it at T. From T, draw a chord to another point A on the circle. Now pick any point B on the far side of the chord, and measure the angle $∠ TB A$ . Pick a second point B and measure again. You get the same angle. That part is not new: angles at the circumference standing on the same arc are equal.

Now measure one more angle: the angle between the tangent and the chord, down at T. It is the same number again.

This should surprise you. The angle at B is made by two chords, BT and BA. The angle at T is made by the chord and a line that lies completely outside the circle, except at one point. The tangent never comes near B. So why do the two angles agree?

This lecture answers that question in three ways: with a picture of B sliding round the circle, with Euclid's proof, and with a second proof from the angle at the centre. By the end, the equal angles will feel forced, not strange.

Naming the parts

The chord TA cuts the circle into two pieces. Each piece is called a segment: the region between the chord and one of the two arcs.

At T, the chord makes an angle with the tangent. Call it $α$ (the Greek letter alpha). In the figure, X is a point on the tangent, so $α = ∠ A TX$ . This angle opens towards one of the two segments. The alternate segment is the other one: the segment on the far side of the chord. "Alternate" here simply means "the other one".

The tangent at T and the chord TA make the angle α. The alternate segment (shaded) is the one across the chord from α. The angle TBA, at any point B on its arc, is also α.

The alternate segment theorem

The angle between a tangent and a chord, at the point of contact, is equal to the angle in the alternate segment.

In the figure: $∠ A TX = ∠ TB A$ for every point B on the arc of the alternate segment.

Read the statement slowly. The angle "in the alternate segment" is an angle at the circumference. Its corner is at a point B on the arc of that segment, and its arms go to the two ends of the chord, T and A. So it is $∠ TB A$ , the angle standing on the chord TA. Other books call it the angle "subtended" by the chord, or the "inscribed angle". The theorem is old. It is Proposition 32 in Book III of Euclid's Elements.

Slide B towards T

Before any proof, here is a way to see why the theorem must be true. You already know that $∠ TB A$ does not change as B moves along its arc. So move B a long way. Move it all the way round, until it is very close to T.

Look at what happens to the two arms of the angle at B. The arm BA becomes almost the chord TA. The arm BT becomes a very short chord near T. A very short chord through T points almost along the tangent. This is the picture from the last lecture: a secant through T turns into the tangent when its second point slides into T.

So when B is almost at T, the angle at B is almost the angle between the tangent and the chord. But the angle at B never changed on the way. So the angle between the tangent and the chord must have been equal to it all along. Try it below.

Explore — slide B round the alternate segment

Choose the angle between the tangent and the chord. Then move B along the shaded arc with the slider or the − / + buttons. Watch the angle at B. Finally, push B as far as it goes, close to T.

Angle at T:

Position of B along the arc: 40%

This figure shows a tangent-chord angle of 70° at T, and three points B₁, B₂ and B₃ on the arc of the alternate segment. Each makes the angle TBA equal to 70°, including B₃, which is close to T. The same is true for every angle at T, including an obtuse one such as 120°, where the alternate segment is the smaller one.

Three points on the arc of the alternate segment all see the chord TA at 70°, the same as the angle between the tangent and the chord.

Whatever angle you choose at T, the readout at B matches it. As B nears T, the short chord BT lies almost along the tangent, and the two angles become one angle. This is not yet a proof, because "almost" is not a reason. But it tells you what to prove, and why it should be true.

Euclid's proof: use the diameter

This is the hard part of the lecture. Spend time here. Every step uses a theorem you already have.

Start with an angle $α = ∠ A TX$ that is less than 90°. Draw the diameter from T, through the centre O, to the opposite point D. Then join D to A.

Euclid's proof. The diameter TD is at 90° to the tangent, so the chord splits that right angle into α and 90° − α. The angle in a semicircle at A is 90°, so the third angle of triangle TAD, at D, is α.

Step 1. $∠ D TX = 9 0^{\circ}$ . The diameter TD contains the radius OT, and the tangent is perpendicular to the radius at the point of contact.
Step 2. The chord TA lies between TD and TX. So it splits the right angle into two parts: $∠ A TX = α$ and $∠ D T A = 9 0^{\circ} - α$ .
Step 3. $∠ T A D = 9 0^{\circ}$ . This is the angle in a semicircle, because TD is a diameter.
Step 4. The angles of triangle TAD add up to 180°. So $∠ T D A = 18 0^{\circ} - 9 0^{\circ} - (9 0^{\circ} - α) = α$ .
Step 5. D is on the arc of the alternate segment. Every other point B on that arc stands on the same chord TA, so $∠ TB A = ∠ T D A = α$ . This is "angles in the same segment are equal".

Look at what just happened. The diameter turns the tangent into something you can use: a right angle. Then the semicircle gives a second right angle. Two right angles in one figure, and the α simply has nowhere else to go.

That is it. That is the whole idea. The tangent-chord angle and the angle at D are the two parts left over when the same 90° is taken away.

A second proof, from the angle at the centre

Here is another route, through the isosceles triangle OTA. It uses the key theorem of the topic: the angle at the centre is twice the angle at the circumference. Press Next to add one step at a time, and read each reason before you go on.

Step through — the proof by the angle at the centre

Press Next to add one step. Read the reason before you go on.

Each step gives one line of the proof and the reason it is allowed. The angle α = ∠ATX is less than 90°, and B is on the arc of the alternate segment.

∠OTX = 90°. The radius OT is perpendicular to the tangent at T.
∠OTA = 90° − α. The chord TA splits the right angle OTX into α and the rest.
∠OAT = 90° − α. OA = OT (both radii), so triangle OTA is isosceles and its base angles are equal.
∠TOA = 180° − 2(90° − α) = 2α. The angles of triangle OTA add up to 180°.
∠TBA = ½ × 2α = α. The angle at the circumference is half the angle at the centre standing on the same arc TA.

So ∠TBA = ∠ATX.

Both proofs lean on the same hinge: the right angle between the tangent and the radius. Without it, nothing in the figure connects the tangent to the rest of the circle.

When the angle at T is obtuse

Both proofs assumed that α is less than 90°. A careful student should ask: what if it is not?

First notice that a chord always makes two angles with the tangent, one on each side. In the next figure, Y and X are on the tangent, on opposite sides of T. The chord makes 60° with TX and 120° with TY. These two angles add up to 180°, because together they make the straight line XTY.

Each angle the chord makes with the tangent matches the angle in the segment across the chord from it: 60° at T matches 60° at B, and 120° at T matches 120° at C.

The acute angle, 60°, is covered by the proofs above. Its alternate segment is the large one, so the angle at B is 60°.

Now take C on the small arc, between T and A. The four points T, C, A and B all lie on the circle, so TCAB is a cyclic quadrilateral. Its opposite angles at B and C add up to 180°. So $∠ TC A = 18 0^{\circ} - 6 0^{\circ} = 12 0^{\circ}$ . That is exactly the angle between the chord and the other part of the tangent, TY.

So the theorem holds for the obtuse angle too. For the 120° angle, the alternate segment is the smaller segment. This is Euclid's second half of the proof, and it is why the alternate segment is not always the major segment. It is always the segment across the chord from the angle you start with.

Stop and think

A chord TA makes an angle of 48° with the tangent at T. A point C is on the small arc, on the same side of the chord as the 48° angle. Is $∠ TC A$ equal to 48°?

Show the answer

No. C is in the same segment as the 48° angle, not the alternate one. The angle at C matches the other tangent-chord angle, $18 0^{\circ} - 4 8^{\circ} = 13 2^{\circ}$ . You can check this with the cyclic quadrilateral: a point B on the far arc sees TA at 48°, and $4 8^{\circ} + 13 2^{\circ} = 18 0^{\circ}$ . Before you use the theorem, always cross the chord.

Stop and think

What does the theorem say when the chord TA is a diameter?

Show the answer

A diameter through T is at 90° to the tangent, on both sides. So the theorem says that the angle at any point B on the circle is 90°. That is the angle in a semicircle. The older theorem is the special case of the new one, which is a good sign that the new one is right.

Worked examples

Worked Example 1 — Both segments

The tangent at T makes an angle of 48° with the chord TA. B is a point on the major arc and C is a point on the minor arc. Find $∠ TB A$ and $∠ TC A$ .

Think first. Which segment is across the chord from the 48° angle? The 48° angle is small, so it opens towards the small segment. Its alternate segment is the large one, where B is.

By the alternate segment theorem, $∠ TB A = 4 8^{\circ}$ .

C is in the other segment. It matches the other tangent-chord angle, $18 0^{\circ} - 4 8^{\circ} = 13 2^{\circ}$ . So $∠ TC A = 13 2^{\circ}$ .

Check. TCAB is a cyclic quadrilateral, so the angles at B and C should add up to 180°: $4 8^{\circ} + 13 2^{\circ} = 18 0^{\circ}$ . ✓

Worked Example 2 — Two chords from the point of contact

XTY is the tangent at T. A and B are points on the circle. $∠ A TX = 5 5^{\circ}$ and $∠ BT Y = 6 5^{\circ}$ . Find the three angles of triangle TAB.

Think first. Each chord from T has its own alternate segment. For the chord TA, the angle at X is across from B. For the chord TB, the angle at Y is across from A.

Chord TA: $∠ TB A = ∠ A TX = 5 5^{\circ}$ (alternate segment theorem).

Chord TB: $∠ T A B = ∠ BT Y = 6 5^{\circ}$ (alternate segment theorem).

Angles in triangle TAB: $∠ A TB = 18 0^{\circ} - 5 5^{\circ} - 6 5^{\circ} = 6 0^{\circ}$ .

Check. The three angles at T on the straight line XTY must add up to 180°: $5 5^{\circ} + 6 0^{\circ} + 6 5^{\circ} = 18 0^{\circ}$ . ✓ Two different reasons give the same answer.

Worked Example 3 — Bringing in the centre

The tangent at T makes an angle of 40° with the chord TA. O is the centre. Find $∠ TO A$ and $∠ OT A$ .

Think first. The theorem gives the angle at the circumference. The angle at the centre is twice that.

A point B on the major arc has $∠ TB A = 4 0^{\circ}$ (alternate segment theorem). So $∠ TO A = 2 \times 4 0^{\circ} = 8 0^{\circ}$ (angle at the centre is twice the angle at the circumference).

Triangle OTA is isosceles, because OT = OA. So $∠ OT A = (18 0^{\circ} - 8 0^{\circ}) \div 2 = 5 0^{\circ}$ .

Check. OT is at 90° to the tangent, and the chord takes 40° of that right angle. So $∠ OT A = 9 0^{\circ} - 4 0^{\circ} = 5 0^{\circ}$ . ✓ This is the step-through proof, run backwards.

Worked Example 4 — Two tangents from a point

Tangents from P touch a circle at A and B. $∠ A PB = 5 0^{\circ}$ . C is a point on the major arc AB. Find $∠ A CB$ .

Think first. PA is a tangent at A, and AB is a chord from A. So the angle between them, $∠ P A B$ , is a tangent-chord angle. Find it first.

PA = PB (tangents from the same point), so triangle PAB is isosceles. Its base angles are $∠ P A B = ∠ PB A = (18 0^{\circ} - 5 0^{\circ}) \div 2 = 6 5^{\circ}$ .

C is across the chord AB from the 65° angle. So $∠ A CB = 6 5^{\circ}$ (alternate segment theorem).

Check. The last lecture showed that $∠ A OB + ∠ A PB = 18 0^{\circ}$ , so $∠ A OB = 13 0^{\circ}$ . Half of that is 65°, the angle at the circumference. ✓

Three traps

Trap 1 — Matching the wrong segment

The equal angle is across the chord, never on the same side as the tangent-chord angle. A point on the same side sees the chord at 180° minus that angle. Before you write the theorem, put your finger on the angle at T, then cross the chord.

Trap 2 — The wrong angle in the triangle

The angle that matches is the one whose corner is on the arc and whose arms go to both ends of the chord. For the chord TA, that is $∠ TB A$ , at B. It is not $∠ BT A$ or $∠ B A T$ . Name the chord first, then find the angle that stands on it.

Trap 3 — A chord that does not start at the point of contact

The theorem needs a tangent and a chord that meet at the point of contact. A chord that crosses the tangent somewhere else, or a line that only looks like a tangent, gives no equal angles. Check both conditions before you quote the theorem.

Try it yourself

Work each one on paper first. Draw a diagram for every question and shade the alternate segment. Open a hint only when you are stuck.

The tangent at T makes an angle of 62° with the chord TA. Find the angle TBA for a point B on the major arc, and the angle TCA for a point C on the minor arc.

Hint 1
Which segment is across the chord from the 62° angle?

Hint 2
B is in the alternate segment of the 62° angle. C matches the other tangent-chord angle, on the far side of T.

Full answer
$∠ TB A = 6 2^{\circ}$ (alternate segment theorem). The other angle between the chord and the tangent is $18 0^{\circ} - 6 2^{\circ} = 11 8^{\circ}$ , and its alternate segment contains C, so $∠ TC A = 11 8^{\circ}$ . Check: TCAB is cyclic, and $6 2^{\circ} + 11 8^{\circ} = 18 0^{\circ}$ . ✓
XTY is the tangent at T, and A and B are on the circle, with A on the X side. $∠ A TX = 7 0^{\circ}$ and $∠ T A B = 4 5^{\circ}$ . Find $∠ BT Y$ and $∠ A TB$ .

Hint 1
The theorem works in both directions. An angle in the triangle can tell you a tangent-chord angle.

Hint 2
$∠ T A B$ stands on the chord TB. Which tangent-chord angle is across that chord from A?

Full answer
$∠ T A B$ is the angle in the alternate segment for the chord TB, so $∠ BT Y = ∠ T A B = 4 5^{\circ}$ . The angles on the straight line XTY add up to 180°, so $∠ A TB = 18 0^{\circ} - 7 0^{\circ} - 4 5^{\circ} = 6 5^{\circ}$ . Check: $∠ TB A = ∠ A TX = 7 0^{\circ}$ (alternate segment for TA), and in triangle TAB, $7 0^{\circ} + 4 5^{\circ} + 6 5^{\circ} = 18 0^{\circ}$ . ✓
The tangent at T is parallel to a chord AB. Prove that TA = TB.

Hint 1
Put X on the tangent on the same side as A. The parallel lines give one pair of equal angles, and the theorem gives another.

Hint 2
$∠ A TX = ∠ T A B$ because they are alternate angles between parallel lines. Now use the alternate segment theorem on $∠ A TX$ .

Full answer
Let X be on the tangent, on the same side as A. Since the tangent is parallel to AB, $∠ A TX = ∠ T A B$ (alternate angles, parallel lines). By the alternate segment theorem, $∠ A TX = ∠ TB A$ , because B is across the chord TA from X. So $∠ T A B = ∠ TB A$ . Two equal angles make triangle TAB isosceles, so TA = TB. Check with numbers: if $∠ A TX = 7 0^{\circ}$ , the base angles are 70° each and $∠ A TB = 4 0^{\circ}$ , and $7 0^{\circ} + 7 0^{\circ} + 4 0^{\circ} = 18 0^{\circ}$ . ✓ Notice that "alternate" appears twice here, with two different meanings.
From a point P outside a circle, a tangent touches the circle at T. A line from P cuts the circle at A and then at B, with PA = 4 cm and AB = 5 cm. Show that triangles PTA and PBT are similar, and use this to find PT.

Hint 1
The two triangles share the angle at P. Find one more pair of equal angles.

Hint 2
$∠ PT A$ is the angle between the tangent and the chord TA. The angle $∠ TB A$ , which is $∠ TBP$ , is across the chord from it.

Full answer
Both triangles contain the angle at P. Also $∠ PT A = ∠ TBP$ , by the alternate segment theorem for the chord TA. Two pairs of equal angles make the triangles similar (the third pair must then match too). Matching sides: PT in the first triangle matches PB in the second, and PA matches PT. So $\frac{PT}{PB} = \frac{P A}{PT}$ , which gives $P T^{2} = P A \times PB$ . Here PB = 4 + 5 = 9 cm, so $P T^{2} = 4 \times 9 = 36$ and PT = 6 cm. Check: the ratio $P A : PT = 4 : 6$ equals $PT : PB = 6 : 9$ , since both are $2 : 3$ . ✓ This result, $P T^{2} = P A \times PB$ , is Euclid's Book III, Proposition 36.

The bigger idea

The alternate segment theorem is not really a new rule. It is the "same segment" theorem with one point pushed to its limit. Slide B round the circle into T, and the chord BT becomes the tangent. The angle stays the same all the way, so the tangent-chord angle must equal it. The proofs make this exact, but the picture is the reason.

This is the same move that the last lecture ended with: a tangent is what a chord becomes when its two ends join. Mathematicians use this move again and again. A result about two points often stays true when the two points come together, and the limit tells you something new. In calculus, the same idea turns the gradient of a chord into the gradient of a curve at one point.

Key concepts in this lesson

The alternate segment is the segment on the far side of the chord from the angle between the tangent and the chord.
The angle between a tangent and a chord at the point of contact equals the angle in the alternate segment, standing on that chord.
Euclid proves it with the diameter through the point of contact, which gives a right angle at the tangent and a right angle in the semicircle.
A second proof uses the isosceles triangle of two radii and the angle at the centre, so both proofs rest on the tangent being perpendicular to the radius.
A chord makes two angles with the tangent, adding up to 180°, and each one matches the segment across the chord from it, which may be the smaller segment.