Same segment and semicircle · Circle geometry

In lesson 1 you saw that $\angle APB$ stays the same wherever P is on the arc. That fact is so useful that it has its own name. This lesson also covers the right angle that hides in every semicircle.

What a segment is

A chord cuts the circle into two pieces. Each piece is called a segment: the region between the chord and one of the arcs. So chord AB makes a major segment (the bigger piece) and a minor segment (the smaller piece).

“P and Q are in the same segment” just means P and Q are on the same side of the chord.

See it for yourself

Angles on the same chordDrag P and Q, then A or B

70°∠APB=70°∠AQB

P and Q are in the same segment (the same side of chord AB), so ∠APB = ∠AQB = 70°. Both stand on the same arc, so both are half of the same angle at the centre.

Slide P and Q along the top arc. The two angles are always equal, whatever you do.
Press Show the angle at the centre. Both angles are half of the same centre angle, so they must be equal. That’s the whole reason.
Move Q to the other side of AB. Now the angles are different, but look at their total.
Press Make AB a diameter. Both angles snap to $90^\circ$ .

Same segment∠APB = ∠AQB

Angle in a semicircleAB a diameter, so ∠APB = 90°

Why the semicircle gives 90°

This is lesson 1 again. When AB is a diameter, A, O and B lie on a straight line, so the angle at the centre is a straight angle, $180^\circ$ . The angle at the circumference is half of that:

\angle APB = \tfrac{1}{2} \times 180^\circ = 90^\circ

Spotting them in a busy diagram

Exam diagrams have lots of lines. These two habits help:

For the same segment, pick a chord and look for two angles whose arms both end at that chord’s endpoints, with both vertices on the same side. They’re usually in a “bow-tie” or “butterfly” shape: two triangles sharing a base, crossing each other.
For the semicircle, look for a line through O that touches the circle at both ends. It’s a diameter, and any triangle standing on it with its third corner on the circle has a right angle at that corner.

A past question, step by step

This one needs both facts, one after the other.

Worked example · JAMB 1994

JAMB 1994 · UME · Q27

In the diagram, $O$ is the centre of the circle and $SOQ$ is a diameter. If $\angle PRS = 38^\circ$ , what is the value of $\angle PSQ$ ?

Which chord do the given angle and the answer share?
We’re given $\angle PRS = 38^\circ$ and asked for $\angle PSQ$ , which is in triangle PQS. We need a way to carry the $38^\circ$ across.
Think first. $\angle PRS$ stands on chord PS. Is there another angle on PS, on the same side, that is in triangle PQS?
Same segment
$\angle PRS$ and $\angle PQS$ both stand on chord PS, and R and Q are on the same side of it:
$\angle PQS = \angle PRS = 38^\circ \quad \text{(angles in the same segment)}$
Think first. SQ goes through O. What does that tell you about the angle at P in triangle PQS?
Semicircle
SOQ is a diameter, so the angle at P, on the circle, is a right angle:
$\angle SPQ = 90^\circ \quad \text{(angle in a semicircle)}$
Finish in triangle PQS
$\angle PSQ = 180^\circ - 90^\circ - 38^\circ = 52^\circ$
The answer is D.

Your turn

JAMB 2003 · UME · Q23

Circle geometry

In the diagram, $O$ is the centre of the circle, $POM$ is a diameter and $\angle MNQ = 42^\circ$ . Calculate $\angle QMP$ .

Worked solution (try it first)

Angles in the same segment are equal.
$\angle MPQ$ and $\angle MNQ$ both stand on arc $MQ$ , so $\angle MPQ = 42^\circ$ .
$PM$ is a diameter, so $\angle MQP = 90^\circ$ (angle in a semicircle).
Triangle $QMP$ : $\angle QMP = 180^\circ - 90^\circ - 42^\circ$
$= 48^\circ$ , option D.

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