Angles in a Circle · Little Blue Book, Visualized

Try this

Keep $P$ above chord $A B$ and drag it from left to right. Does the reading for $∠ A P B$ change?
Drag $P$ past $A$ or $B$ to the other side of the chord. What do the new reading and the old one add up to?
The blue angle is the central angle $∠ A O B$ , standing on the same arc. How is it related to the red reading? Drag $A$ and $B$ to change the chord: does the relationship survive?
Drag $A$ and $B$ so that chord $A B$ passes through the center $O$ . What is the inscribed angle now?

An angle whose vertex is on the circle and whose two sides both cut the circle is an inscribed angle. Put the four observations together and you get:

∠ A P B = \frac{1}{2} ∠ A O B

Here $∠ A O B$ means the measure of the arc that $∠ A P B$ stands on. Same arc, same inscribed angle.

Predict first

Chord $A B$ is fixed. With $P$ on the major arc, $∠ A P B = 5 0^{\circ}$ . Drag $P$ onto the minor arc. What is $∠ A P B$ now, in degrees?

Choose an answer first; then you can see the result.

Why

The whole proof rests on one fact: all radii are equal, so the center and two points on the circle always form an isosceles triangle.

Proof: an inscribed angle is half the central angle

Step 1 / 6

What to prove
$P$ is on the circle; we want $∠ A P B = \frac{1}{2} ∠ A O B$ . The figure shows $6 5^{\circ}$ and $13 0^{\circ}$ right now, but we need a reason that works for every position.

$∠ A P B = \frac{1}{2} ∠ A O B ?$
Add a diameter
Left side: an isosceles triangle
Same on the right; add up
When the center is outside the angle
Conclusion

If the vertex leaves the circle

Break it

Counterexample lab: remove one condition

Conclusion: $∠ A P B$ is half of arc $A B$

While the vertex is on the circle, the angle is half the arc however you drag.

Practice

Exercise 1Basic

In circle $O$ the minor arc $A B$ has central angle $∠ A O B = 11 0^{\circ}$ , and point $P$ is on the major arc $A B$ . Find $∠ A P B$ in degrees.

Exercise 2Basic

$A B$ is a diameter of a circle, $C$ is a point on the circle, and $∠ C A B = 3 8^{\circ}$ . Find $∠ C B A$ in degrees.

Exercise 3Advanced

A circle has radius $2$ and chord $A B = 22$ . What can the inscribed angle on chord $A B$ be, in degrees? (List every possibility, separated by commas.)

Exercise 4Challenge

Four points $A$ , $B$ , $C$ , $D$ lie on a circle in that order. Chords $A C$ and $B D$ meet at a point $E$ inside the circle. Arc $A B$ is $7 0^{\circ}$ and arc $C D$ is $5 0^{\circ}$ . Find $∠ A E B$ in degrees.

Mistake Museum · Exhibits from this stop

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Exhibit

Thinking a chord has only one inscribed angle

The circle has radius

4

and chord

A B = 4

, so the inscribed angle on chord

A B

3 0^{\circ}

What went wrong? Think first, then open

Diagnosis Chord $A B$ splits the circle into two arcs. With the vertex on the major arc the inscribed angle is $3 0^{\circ}$ ; with the vertex on the minor arc it is $15 0^{\circ}$ . The problem says “the inscribed angle on the chord” without saying which side the vertex is on, so you need both.

Find the central angle first:

△ O A B

is equilateral, so

∠ A O B = 6 0^{\circ}

. On the major arc the inscribed angle is

3 0^{\circ}

; on the minor arc it is

18 0^{\circ} - 3 0^{\circ} = 15 0^{\circ}

. The answer is

3 0^{\circ}

15 0^{\circ}

“An arc gives one inscribed angle; a chord gives two”

Exhibit

Treating an angle as inscribed when its vertex is not on the circle

Chords

A C

and

B D

meet at a point

E

inside the circle, and arc

A B

8 0^{\circ}

, so

∠ A E B = 4 0^{\circ}

What went wrong? Think first, then open

Diagnosis “Half the arc it stands on” only holds when the vertex is on the circle. $E$ is inside, so $∠ A E B$ is bigger than $4 0^{\circ}$ : you also have to add half of the opposite arc $C D$ .

Join

B C

∠ A E B

is an exterior angle of

△ E B C

, so it equals

∠ A C B + ∠ D B C

, that is, half of arc

A B

plus half of arc

C D

“First check whether the vertex is on, inside or outside the circle, then pick the formula”