GCSE · Maths · OCR · Spec J560

Alternate segment theorem

Two angles: one made by a tangent, one tucked into the far side of a chord. Slide the third point wherever you like and they refuse to differ. Why?

Maths · Circle theorems

Drag P round the top. What refuses to change?

Tangent–chord angle at A 40.0°. Angle APB at P 40.0°. PA 10.3 cm. PB 10.3 cm. Classification: The claim to test. Relationship: While P is on the far side of AB from T, angle APB equals the angle between the tangent AT and the chord AB.

Tangent–chord angle at A40.0°Angle APB at P40.0°PA10.3 cmPB10.3 cmTry P close to A, then close to B.

The claim to test

While P is on the far side of AB from T, angle APB equals the angle between the tangent AT and the chord AB.

A is where the tangent touches the circle, and AB is a chord. Keep P on the opposite side of AB from T. If you carry P across below AB, the match stops: the theorem is about the far side.

Maths · Algebra

From “it keeps working” to “it always works”

A theorem is a statement shown to be always true by reasoning from facts we already trust. Measuring can't do that. This can: every line below leans on one earlier circle fact.

GoalShow that the tangent–chord angle TAB equals angle APB, whatever size angle APB is

∠APB = x

name it

Call the angle at P by a letter, not a number. Now nothing we do depends on it being 40°, or any size in particular.

Step 1 of 7

Call the angle at P by a letter, not a number. Now nothing we do depends on it being 40°, or any size in particular.

Watch out: This is the case drawn on the board: angle APB is acute, so the centre O sits on the same side of AB as P.

Predict, then check

Picture triangle PQR inside a circle with all three corners on the circle.

A tangent touches the circle at corner P. Outside the triangle, the angle between that tangent and side PQ is marked. Which interior angle of the triangle does it equal?

Which idea sounds right to you?

Why are the two angles equal?

A tangent touches a circle at A. AB is a chord and P is a point on the circle on the far side of AB. The angle between the tangent and AB equals angle APB.

Which explanation is closest to what you think?

They are alternate angles, and alternate angles are equal.It only works if the tangent is parallel to one of the chords.It is the alternate segment theorem: the tangent–chord angle equals the angle in the alternate segment.Angle APB would match the tangent angle wherever P is, on either side of AB.

How sure are you?

I'm sureI think soI'm guessing

They are alternate angles, and alternate angles are equal.

This is a common misconception

The name is the trap. Alternate angles are equal only between parallel lines (the Z shape). Look back at the board: there is no pair of parallel lines in the figure. “Alternate” in this theorem points at the alternate segment, the far side of the chord, not at a Z.

It only works if the tangent is parallel to one of the chords.

This is a common misconception

No parallel line is needed. Drag P anywhere on the far side of AB and the angles still match, whether or not any chord happens to be parallel to the tangent. The match comes from the chord, not from parallel lines.

It is the alternate segment theorem: the tangent–chord angle equals the angle in the alternate segment.

This matches the evidence

Yes, and that is the exact reason to write in an answer. The tangent–chord angle is against chord AB, and angle APB stands on that same chord from the far side.

Angle APB would match the tangent angle wherever P is, on either side of AB.

This is a common misconception

Close, but the side matters. Carry P below AB on the board and the two readouts stop agreeing. The theorem is about the alternate segment, the side of the chord away from your tangent–chord angle.

Write “alternate segment theorem”, not “alternate angles”. The match works because of the chord, with no parallel lines needed.

Worked problem: a chain of angles

Problem

Triangle ABC is drawn in a circle, with BC a diameter. TS is the tangent at A: T is nearer B, S is nearer C. Angle TAB = 35°. Find angle SAC, with a reason for each step.

Mark this answer

Where does this answer go wrong?

D, E and F are points on a circle, and MN is the tangent at D. M is towards E and N is towards F. Angle MDE = 52° and angle EDF = 67°. Find angle DEF and angle DFE.

A student's answer — which line goes wrong?

Label 52° between the tangent and DE, and 67° at D inside the triangle.
The 52° is between the tangent and a chord, so by the alternate segment theorem angle DEF = 52°.
Angles in triangle DEF add to 180°, so angle DFE = 180° − 67° − 52° = 61°.
So angle DEF = 52° and angle DFE = 61°.

WHAT YOU'VE LEARNED

A quick recap of today's lesson.

Slide a point round a circle and watch two angles refuse to differ.

What you need to know

A chord drawn from the point where a tangent touches a circle makes an angle with the tangent: the tangent–chord angle.
Alternate segment theorem: the tangent–chord angle equals the angle at the circumference standing on the same chord, in the alternate segment.
The alternate segment is the one on the far side of the chord from your tangent–chord angle.
Have a goYou have a tangent–chord angle in front of you. Does its matching angle live on the same side of the chord, or the far side?
The far side: the alternate segment.
The alternate segment is the one on the opposite side of the chord from your angle, so the same side is the wrong segment.
To find the match, name the chord the angle is against. The match stands on that chord, from the far side.
Have a goJo bets you a biscuit: the tangent–chord angle against chord PQ is 38°, R sits somewhere on the far side of PQ, and you must name angle PRQ with no protractor. Take the bet?
Yes. Angle PRQ = 38°.
R is in the alternate segment of chord PQ, so the angle at R standing on PQ equals the tangent–chord angle, wherever R sits on that side.
Dragging a point and measuring only checks particular cases. It cannot show the theorem is true for every position.
The proof borrows three earlier facts: centre angle is double, radii make an isosceles triangle, tangent meets radius at 90°.
Have a goThe angle at the circumference is 35°, so the angle at the centre on the same chord is 70°. The two radii make an isosceles triangle with 70° at the top. How big is each base angle?
55°
The angles of the triangle add to 180°, so the two equal base angles share 180° − 70° = 110°, which is 55° each.
With a tangent at one corner of an inscribed triangle, each outside angle matches the interior angle at the opposite corner.
“Alternate” here is not alternate angles: those need parallel lines, and these figures usually have none. Quote “alternate segment theorem”.
In angle chains, give a reason on every line and add each angle you find to the diagram.

The big picture

The angle between a tangent and a chord equals the angle at the circumference in the alternate segment, standing on that same chord. You find the pair by naming the chord first. It is a theorem, not just a pattern, because a short proof works for any size of angle. And “alternate” means the far-side segment, not alternate angles on parallel lines.

Key points

1The tangent–chord angle equals the angle at the circumference in the alternate segment, standing on the same chord.

2The alternate segment is the far side of the chord. It has nothing to do with alternate angles or parallel lines.

3Proof chain: angle at the centre 2x, base angles 90° − x, so the tangent–chord angle is 90° − (90° − x) = x.

4In an answer, write “alternate segment theorem” as the reason and add every angle you find to the diagram.

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Worked example

Problem

A tangent touches a circle at A. P is on the circle, on the far side of chord AB, and PA = PB. The tangent makes 64° with AB. Find angle PAB.

⚠ Watch out

Pairing the tangent–chord angle with an angle at an end of its chord. The match stands on the chord, at the circumference, on the far side.

🧠

Memory hook

Chord first, then jump: find the chord your angle is against, then hop across it to the circumference.

✓

Check yourself

Sketch from memory: circle, chord AB, tangent at A, point P on the far side. Mark the two equal angles, then list the three earlier facts the proof borrows.

Flashcards

(13)

What is the tangent–chord angle?

The angle between a tangent and a chord that starts at the point where the tangent touches the circle.

Alternate segment theorem: what does it say?

The tangent–chord angle equals the angle at the circumference in the alternate segment, standing on the same chord.

What is the “alternate segment”?

The segment on the far side of the chord from the tangent–chord angle you started with.

How do you find the angle that matches a tangent–chord angle?

Name the chord the angle is against. The match is the angle that stands on that chord, at the circumference, on the far side.

Twenty diagrams in a row show equal angles. Is the theorem proved?

No. Measuring covers particular cases only. A proof has to work for every position.

What is a theorem?

A statement shown to be always true by reasoning from facts we already trust.

In the proof, the angle at the circumference is x. What is the angle at the centre on the same chord?

2x: the angle at the centre is twice the angle at the circumference.

Why is triangle OAB isosceles in the proof?

OA and OB are both radii, so they are equal and the base angles are equal.

In the proof, how big is each base angle of triangle OAB?

90° − x: share (180° − 2x) between the two equal angles.

How does the proof use the tangent?

A tangent meets the radius at 90°, so ∠TAB = 90° − (90° − x) = x.

Tangent at corner P of an inscribed triangle PQR. The outside angle against side PR equals which interior angle?

∠PQR, the corner opposite side PR.

A student gives “alternate angles” as the reason. What is wrong?

Alternate angles need parallel lines, and these figures usually have none. The reason is the alternate segment theorem.

What two habits keep a chain of angles safe?

A reason on every line, and every angle you find added to the diagram.

Tap any card to flip it, or use Study as deck to go through them one at a time. In the full lesson these run as a spaced-repetition deck — you rate each card Hard, Good or Easy and the tricky ones keep coming back until they stick.

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From “it keeps working” to “it always works”

Why are the two angles equal?

Where does this answer go wrong?

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