Proving the inscribed angle theorem

When you reach for it

Inscribed angles states the fact: an angle whose vertex is on a circle and whose sides are chords measures half the arc it opens onto. Measure a few and it is easy to believe. Proving it is harder than it looks, because the one argument that works cleanly needs the center of the circle to sit on a side of the angle, and in most drawings it does not.

So the proof is a proof by cases, and this lesson is about writing it. The plan fits in one sentence: handle the angle with the center on a side first, then reach every other angle by adding two of those or subtracting one from another. Once it is written, “Inscribed Angle Theorem” can stand in the Reasons column of a two-column proof, and the rest of proofs about circles leans on it: right angles in semicircles, equal angles on one arc, cyclic quadrilaterals, and the angles made by chords and secants.

You reach for the same pattern again whenever a fact about circles is easy in one special position. Prove it there, then draw an auxiliary diameter to cut the general figure into special pieces.

The first big use of the theorem is also proved here in full. A cyclic quadrilateral is one whose four vertices lie on one circle. Draw a few and measure: the angles at opposite corners always add to 180^\circ. You reach for that fact when four points of a figure are known to lie on a circle and you need an angle. Each angle of the quadrilateral is an inscribed angle, so its proof is one citation of the Inscribed Angle Theorem, used twice.

You reach for its converse the other way round: when two opposite angles are known to be supplementary and you want the four corners on one circle. That is how you prove that every rectangle and every isosceles trapezoid fits on a circle, and it is the first step of many harder proofs, which put four points on a circle so that the inscribed angle theorem can be used on them.

The procedure

The words. \angle APB is inscribed in a circle when its vertex P is on the circle and its sides meet the circle again at A and B, so that \overline{PA} and \overline{PB} are chords. The angle intercepts the arc \overset{\frown}{AB} whose points, apart from A and B, lie inside the angle. That is the arc from A to B that does not pass through P.

The measure of an arc comes from central angles. A minor arc measures the same as the central angle that cuts it off, a semicircle measures 180^\circ, and a major arc measures 360^\circ minus its minor arc. Arcs that meet end to end add, by arc addition.

The theorem.

Given. \angle APB is inscribed in a circle with center O and intercepts \overset{\frown}{AB}.

Prove. m\angle APB = \tfrac{1}{2}\, m\overset{\frown}{AB}.

Plan. The center O lies somewhere in the plane of the angle, and there are only three places it can be: on one of the sides, inside the angle, or outside it (Figure 1). Prove each place separately. The first case uses an isosceles triangle and an exterior angle. The other two draw the diameter from P and use the first case twice.

The cases leave nothing out. The inside of the angle, the angle itself, and the outside split the plane with no overlap, so O is in exactly one of them. If O is on the angle, it is on \overline{PA} or on \overline{PB} and not on both, since the two sides share only P and O is not on the circle. The case with O on \overline{PB} is the case with O on \overline{PA} with the letters A and B swapped, so it needs no proof of its own.

Case 1: the center on a side

Given. \overline{PA} is a diameter, and B is another point of the circle.

Prove. m\angle APB = \tfrac{1}{2}\, m\overset{\frown}{AB}.

Plan. Draw the radius \overline{OB}. Then \triangle OPB has two radii for sides, so its angles at P and B are equal; call each x. The central angle \angle AOB sits outside that triangle, next to a straight line, so it is the sum of the two far angles, 2x. And the arc is the central angle.

Line 1 is what the whole case rests on. Because P, O and A are on one line, \angle AOB is an exterior angle of \triangle OPB, and line 5 is allowed. It is also why line 7 may call \overset{\frown}{AB} a minor arc: the inscribed angle’s arc is the one away from P, and P is straight behind O. Lines 3 and 5 cite theorems proved earlier, in proving base angles congruent and proving the exterior angle theorem, and nothing about inscribed angles, so this case stands on its own.

Case 2: the center inside the angle

Given. \angle APB is inscribed, and O is inside it.

Prove. m\angle APB = \tfrac{1}{2}\, m\overset{\frown}{AB}.

Plan. Draw the diameter from P. It passes through O, which is inside the angle, so it splits \angle APB into \angle 1 and \angle 2, and it ends at a point D that splits the arc into \overset{\frown}{AD} and \overset{\frown}{DB}. Each small angle has the center on a side, which is Case 1. Add.

Line 3 carries two position facts, and each is used once. The ray inside the angle is what angle addition needs in line 4; the point on the arc is what arc addition needs in line 5. Both are read from where O is, which is why the case split is made on O and nothing else. The arcs in lines 6 and 7 are the right ones: \overset{\frown}{AD} is the part of \overset{\frown}{AB} inside \angle APD, and it does not pass through P.

Case 3: the center outside the angle

Given. \angle APB is inscribed, and O is outside it.

Prove. m\angle APB = \tfrac{1}{2}\, m\overset{\frown}{AB}.

Plan. Draw the diameter \overline{PD} again. Now it misses the angle, and both sides of the angle lie on the same side of it, one nearer than the other. Name the points so that B is the nearer one, as in Figure 4; the letters are ours to choose. Then \angle APD and \angle BPD are both Case 1 angles, and \angle APB is the first with the second taken away.

Line 3 is a position fact, the kind a figure may be read for, but Figure 4 is one drawing and the case covers every angle with the center outside, so here is why it always holds. \triangle OPA has two radii for sides, so its base angles are equal, and two equal angles of a triangle are each less than 90^\circ. Since O is on \overrightarrow{PD}, that says m\angle APD < 90^\circ, and in the same way m\angle BPD < 90^\circ. So \overrightarrow{PA}, \overrightarrow{PB} and \overrightarrow{PD} all point into one half-plane, D’s side of the line through P perpendicular to \overline{PD}, and of three rays from one point into one half-plane, one lies between the other two. It is not \overrightarrow{PD}, because O is on it and O is outside \angle APB. So one of \overrightarrow{PA} and \overrightarrow{PB} lies between \overrightarrow{PD} and the other, and that one is the ray we named \overrightarrow{PB}. Then B is a point of the circle inside \angle APD, which is what it means to be on \overset{\frown}{AD}.

That completes the proof. Cases 2 and 3 cite Case 1, which was proved without the theorem, so nothing is used before it is proved.

Quoting the theorem afterwards. From now on “Inscribed Angle Theorem” is a reason, and it needs no case: every inscribed angle is covered. Its if-part is that the vertex is on the circle and both sides are chords. Name the arc as the one inside the angle, the arc that does not pass through the vertex. Two consequences are used so often they have their own names, the angle in a semicircle (Example 4) and inscribed angles on the same arc (Example 5). A third, the opposite angles of a cyclic quadrilateral, is proved next, together with its converse.

Proving opposite angles of a cyclic quadrilateral

The words. Quadrilateral ABCD is inscribed in a circle, or cyclic, when A, B, C and D all lie on the circle. Here they lie on it in that order, going round, so the sides \overline{AB}, \overline{BC}, \overline{CD}, \overline{DA} are chords that do not cross. Its opposite angles are \angle DAB with \angle BCD, and \angle ABC with \angle CDA. When no other angle could be meant, \angle A is short for \angle DAB.

Two points cut a circle into two arcs, so a two-letter name such as \overset{\frown}{BD} can be ambiguous. A three-letter name settles it: \overgroup{BCD}, with the arc drawn over all three letters, is the arc from B to D that passes through C. A two-letter name keeps the meaning fixed above: the arc an inscribed angle intercepts, the one that does not pass through its vertex.

Given. A, B, C and D lie on a circle in that order.

Prove. m\angle DAB + m\angle BCD = 180^\circ and m\angle ABC + m\angle CDA = 180^\circ.

Plan. \angle DAB and \angle BCD are both inscribed. Between them they open onto the two arcs that B and D cut the circle into (Figure 5), and those two arcs make the whole circle, 360^\circ. Each angle is half its arc, so together they are half of 360^\circ.

The other pair needs no new idea. Run lines 2 to 10 again with B and D in the places of A and C: \angle ABC intercepts \overgroup{CDA}, \angle CDA intercepts \overgroup{ABC}, and those two arcs make the whole circle. Or use the angle sum of a quadrilateral: the four angles add to 360^\circ, one pair takes 180^\circ, so the other pair has the 180^\circ that is left.

Line 2 is the only place the order of the vertices is used, and it is used hard. The arc an angle opens onto is fixed by where the other two vertices are; if the four points were joined out of order, the angle at A would open onto a different arc (Exercise 38). Notice also what the proof never mentions: where the center is. The Inscribed Angle Theorem covers every inscribed angle, so nothing here splits into cases. That is the three cases above paying off. From here on the result is quoted as the cyclic quadrilateral theorem.

The converse: supplementary opposite angles put a quadrilateral on a circle

The converse of the cyclic quadrilateral theorem swaps its if-part and its then-part, and a converse is a new claim that needs its own proof. Here it is true, with one condition added.

Given. ABCD is a convex quadrilateral, its vertices named in order round it, with m\angle DAB + m\angle BCD = 180^\circ.

Prove. A, B, C and D lie on one circle.

Plan. Any three points not on one line lie on exactly one circle, the circumscribed circle of the triangle they make. So draw the circle through A, B and D. The question is only whether C is on it. Follow the diagonal from A through C until it meets the circle again at a point C'. The quadrilateral ABC'D is cyclic, so the theorem gives it supplementary opposite angles, and that forces \angle BC'D to be the same size as \angle BCD. Then show that if C were anywhere else on that line than at C', the two angles could not be the same size. This last step is an indirect proof.

Why line 9 holds. Suppose C is not C'. Both lie on \overrightarrow{EC} (line 5), so one of them is between E and the other. That gives two possibilities, the two panels of Figure 6.

C between E and C', so C is inside the circle. In \triangle BCC' the side \overline{C'C} runs on past C to E, so \angle BCE is an exterior angle, and by the Exterior Angle Theorem m\angle BCE = m\angle CBC' + m\angle BC'C. The angle at B is more than 0^\circ, so m\angle BCE > m\angle BC'C, and \angle BC'C is \angle BC'E because C is on \overrightarrow{C'E}. So m\angle BCE > m\angle BC'E. The same argument in \triangle DCC' gives m\angle ECD > m\angle EC'D. Now E is between B and D, so \overrightarrow{CE} is inside \angle BCD and \overrightarrow{C'E} is inside \angle BC'D, and the angle addition postulate adds the pieces:

That contradicts line 8.

C' between E and C, so C is outside the circle. The same argument with the letters C and C' swapped gives m\angle BC'D > m\angle BCD, which also contradicts line 8.

Both possibilities are impossible, so C is C', and C is on the circle. That completes the proof.

Where “convex” was used. Line 3. The diagonals of a convex quadrilateral cross, which puts C on the far side of \overline{BD} from A and lets the diagonal from A carry C to the arc opposite A. Without it the converse is false, as the dart in Boundaries shows.

Quoting both afterwards. Once proved, the theorem and its converse can be combined into one biconditional: a convex quadrilateral is cyclic if and only if one pair of its opposite angles is supplementary. One pair is enough, because the angle sum of 360^\circ then makes the other pair supplementary too. In a proof, cite the theorem to get angles from a circle and the converse to get a circle from angles.

Worked examples

Boundaries

The vertex must be on the circle. Move it inside and the angle is half the sum of two arcs (Example 6); move it outside and it is half their difference, as in angles with a vertex outside the circle and Exercise 28. Neither is an inscribed angle, and the theorem says nothing about them until a proof like Example 6 connects them.

The arc is the one inside the angle. An inscribed angle less than 90^\circ opens onto a minor arc, a right angle onto a semicircle, and an angle more than 90^\circ onto a major arc. The version “half the central angle” holds only for the first kind (Example 2).

Each case needs its own position facts. Case 1 needs P, O and A on one line; Case 2 needs the diameter to pass inside the angle; Case 3 needs one side between the diameter and the other. A proof that does only Case 1 has proved the theorem only for angles with a diameter as a side.

The sides must both be chords. If one side of the angle is tangent to the circle at P, the angle is not inscribed, though it too measures half its arc. That is the tangent chord angle, which needs the tangent to be perpendicular to the radius. Some books run the order the other way: they prove the tangent chord angle first, then get every inscribed angle as the difference of two tangent chord angles at P, with no cases at all. Either order is sound; one proof may not mix them.

A cyclic quadrilateral’s vertices must be joined in order round the circle. Four points on a circle can also be joined into a crossed figure, with two sides that cut across each other. Its angles at “opposite” corners open onto the same arc, so they are congruent, not supplementary (Exercise 38). Every statement of the cyclic quadrilateral theorem carries the order in its Given, as line 1 of its proof does.

The converse needs convex. Take a dart ABCD, a concave kite: m\angle A = 30^\circ, m\angle B = m\angle D = 60^\circ, and the inside angle at C is 210^\circ, a reflex angle. The angle \angle BCD between the rays \overrightarrow{CB} and \overrightarrow{CD} is the other side of that corner, 360^\circ - 210^\circ = 150^\circ, so m\angle A + m\angle BCD = 30^\circ + 150^\circ = 180^\circ. Yet ABCD is not cyclic: C lies inside \triangle ABD, so it is inside the circle through A, B and D. The proof of the converse breaks at line 3, since the diagonals of a dart do not cross.

Supplementary is about opposite angles only. Neighboring angles of a cyclic quadrilateral can be anything that fits, and two neighboring angles adding to 180^\circ says nothing about a circle: every parallelogram has that, and Example 11 (b) is not cyclic.

Flat geometry only. Case 1 cites the Exterior Angle Theorem, which rests on the angle sum of a triangle and so on the parallel postulate. On a sphere, where a triangle’s angles add to more than 180^\circ, the theorem fails, and the cyclic quadrilateral theorem, which rests on it, fails with it.

Where this goes next

With the theorem as a reason, a family of results comes in a few lines each: the angle in a semicircle is a right angle, angles on the same arc are equal, and the opposite angles of a cyclic quadrilateral add to 180^\circ, proved above together with its converse.

Joined to the Exterior Angle Theorem, it measures every angle two lines can make across a circle: two chords crossing inside (Example 6), and two secants or tangents meeting outside (Exercise 28). All of inscribed angles as a working tool starts from the proof in this lesson.

The cyclic quadrilateral theorem and its converse together sort the family of quadrilaterals: every rectangle, square and isosceles trapezoid is cyclic; a parallelogram is cyclic only when it is a rectangle, a rhombus only when it is a square, and a trapezoid only when it is isosceles or a rectangle (Exercises 29, 30 and 39). That is the quadrilateral case of inscribed polygons.

The converse is also a tool for proofs that start with no circle at all. Find two opposite right angles, or any supplementary pair, and a circle appears through four points of the figure; then every inscribed angle fact applies to them. Exercise 31 does this with the two tangents from an outside point, whose right angles come from the tangent and the radius. Drawing such a circle on purpose is a standard move in harder geometry, used again and again in proofs about circles.

Exercises

Reading the proofs

1. Line 3 of Case 1 cites the Base Angles Theorem for \triangle OPB. Which two sides of the triangle are equal, and why?
2. Line 5 of Case 1 uses line 1, that O is between P and A. Why is that needed, and what would \angle AOB be to \triangle OPB if P, O and A were not on one line?
3. Line 3 of Case 2 says two things: \overrightarrow{PD} is inside \angle APB, and D is on \overset{\frown}{AB}. Which later line uses each, and what would go wrong without it?
4. Case 2 adds two Case 1 angles, and Case 3 subtracts one from another. What about the position of O makes the difference?
5. Line 2 of the proof that opposite angles of a cyclic quadrilateral are supplementary says \angle DAB intercepts \overgroup{BCD}. Why not \overgroup{BAD}?
6. Line 6 of that proof says the two arcs add to 360^\circ. Explain why, whether or not \overline{BD} is a diameter.
7. In the proof of the converse, which line uses the hypothesis that ABCD is convex, and what does that line give the rest of the proof?

Numbers through the cases

In Exercises 8, 9 and 14, \overline{PA} is a diameter of a circle with center O, as in Figure 2. In Exercises 10 to 13, \overline{PD} is a diameter and the points are arranged as in Figure 3 (for 10 and 11) or Figure 4 (for 12 and 13). Each arc is the one that does not pass through P.

8. m\angle APB = 27^\circ. Find m\angle AOB, m\angle POB and m\overset{\frown}{PB}, the arc not passing through A.
9. m\overset{\frown}{AB} = 88^\circ. Find m\angle APB, m\angle OBP and m\angle POB.
10. m\angle APD = 31^\circ and m\angle DPB = 38^\circ. Find m\overset{\frown}{AD}, m\overset{\frown}{DB}, m\overset{\frown}{AB} and m\angle APB.
11. m\overset{\frown}{AD} = 124^\circ and m\overset{\frown}{DB} = 96^\circ. Find m\angle APB and m\overset{\frown}{AB}, and the measure of the angle \angle AOB between the radii to A and B.
12. m\angle APD = 42^\circ and m\angle BPD = 17^\circ. Find m\overset{\frown}{AB} and m\angle APB.
13. m\overset{\frown}{AD} = 150^\circ and m\overset{\frown}{AB} = 64^\circ. Find m\angle BPD and m\angle APB.
14. m\angle OPB = (3x + 4)^\circ and m\angle AOB = (7x - 10)^\circ. Find x and both angles.
15. \triangle ABC is inscribed in a circle. The arc \overset{\frown}{BC} not passing through A measures (2x + 10)^\circ, the arc \overset{\frown}{CA} not passing through B measures (3x)^\circ, and the arc \overset{\frown}{AB} not passing through C measures (4x - 10)^\circ. Find x and the three angles of the triangle.
16. The inscribed angle \angle APB and the central angle \angle AOB both open onto the same minor arc \overset{\frown}{AB}, and m\angle AOB - m\angle APB = 37^\circ. Find both angles.

Cyclic quadrilaterals: angles and arcs

17. ABCD is inscribed in a circle, m\angle A = 58^\circ and m\angle B = 121^\circ. Find m\angle C and m\angle D.
18. PQRS is inscribed in a circle, with the vertices in that order, and m\angle P = 3\,m\angle R. Find m\angle P and m\angle R.
19. ABCD is inscribed in a circle, m\angle A = {(5x - 8)^\circ} and m\angle C = {(3x + 12)^\circ}. Find x and both angles.
20. ABCD is inscribed in a circle with m\overset{\frown}{AB} = 70^\circ, m\overset{\frown}{BC} = 100^\circ, m\overset{\frown}{CD} = 110^\circ and m\overset{\frown}{DA} = 80^\circ, each arc passing through neither of the other two vertices. Find all four angles of ABCD.
21. ABCD is inscribed in a circle, m\angle A = 100^\circ, and the arc from A to B that passes through neither C nor D measures 76^\circ. Find m\overgroup{BCD}, m\overgroup{DAB}, m\angle C, and the measure of the arc from D to A that passes through neither B nor C.
22. ABCD is inscribed in a circle, C is between B and X, and m\angle DCX = 71^\circ. Find m\angle DAB and m\angle BCD.

Is it cyclic?

23. Each list gives the angles of a convex quadrilateral in the order \angle A, \angle B, \angle C, \angle D. Which are cyclic? (a) 88^\circ, 97^\circ, 92^\circ, 83^\circ. (b) 105^\circ, 75^\circ, 105^\circ, 75^\circ. (c) 60^\circ, 100^\circ, 120^\circ, 80^\circ.
24. Convex quadrilateral ABCD has m\angle A = (4x + 5)^\circ and m\angle C = (6x - 25)^\circ. For which x is ABCD cyclic? Give both angles for that x.
25. ABCD is inscribed in a circle, with the vertices in that order, m\angle A = 2\,m\angle B and m\angle D = m\angle C + 20^\circ. Find all four angles.

Writing proofs

26. Rewrite the proof of Case 1 without the Exterior Angle Theorem. Use the Triangle Angle Sum Theorem for \triangle OPB and the linear pair \angle POB, \angle AOB instead.
27. Given. The inscribed angle \angle APB intercepts \overset{\frown}{AB}, the inscribed angle \angle CQD intercepts \overset{\frown}{CD}, and m\overset{\frown}{AB} = m\overset{\frown}{CD}. Prove. \angle APB \cong \angle CQD.
28. Given. P is outside a circle; A, B, C, D are on the circle; A is between P and B, and C is between P and D. Prove. m\angle BPD = \tfrac{1}{2}\left(m\overset{\frown}{BD} - m\overset{\frown}{AC}\right), where \overset{\frown}{BD} does not pass through A and \overset{\frown}{AC} does not pass through D. (Draw \overline{AD}.)
29. Given. Rhombus ABCD is inscribed in a circle, with the vertices in that order. Prove. ABCD is a square.
30. Given. Trapezoid ABCD, with \overline{AB} \parallel \overline{DC}, is inscribed in a circle, with the vertices in that order. Prove. \angle DAB \cong \angle ABC.
31. Given. P is outside a circle with center O; \overline{PA} and \overline{PB} are tangent to the circle at A and B; quadrilateral PAOB is convex. Prove. P, A, O and B lie on one circle, and m\angle APB + m\angle AOB = 180^\circ.
32. Given. ABCD is inscribed in a circle, with the vertices in that order, and \angle DAB \cong \angle BCD. Prove. Both angles are right angles, and \overline{BD} is a diameter.

Reasoning

33. Explain why every inscribed angle falls under exactly one of the three cases.
34. An inscribed angle measures 100^\circ. Which case can it be in? Explain.
35. Example 2 shows that “an inscribed angle is half the central angle on the same arc” can fail. For exactly which inscribed angles does it hold?
36. Case 2 cites Case 1 twice. Explain why that is not arguing in a circle, and why Case 1 itself could not be proved the way Case 2 is, by drawing the diameter from P.
37. The converse asks for only one pair of opposite angles to be supplementary. Explain why the other pair then comes free.
38. A, B, C, D lie on a circle in that order. Join them instead in the order A, C, B, D, making a crossed figure with sides \overline{AC}, \overline{CB}, \overline{BD}, \overline{DA}. Show that its angles at A and at B, which are \angle DAC and \angle CBD, are congruent, and explain why that does not contradict the cyclic quadrilateral theorem.
39. Which of these are always cyclic: a rectangle, a square, a rhombus, a kite? For each one that is not, say exactly when it is.
40. The proof of the converse starts with a circle through A, B and D only. Why could it not start with a circle through all four vertices, and why is there always a circle through those three?