Circle theorems link the geometry of a circle to angles. The angle in a semicircle theorem states that any angle inscribed in a semicircle (with the diameter as the base) is always a right angle (90°). The angles in the same segment theorem states that angles subtended by the same arc (from the same part of the circle) are equal. A cyclic quadrilateral is a quadrilateral inscribed in a circle, and its opposite angles sum to 180°.
The angle in a semicircle is always 90°
A triangle is inscribed in a semicircle with the diameter as one side. What is the angle opposite the diameter?
TheoremAngle in a semicircle = 90°WhyThe angle subtended by a diameter at any point on the circle is always a right angle
This works for any point on the semicircle. The diameter is always the longest side (hypotenuse), and the angle opposite it is 90°.
Angles in the same segment are equal
Two angles are inscribed in a circle and subtend the same arc. Are they equal?
TheoremAngles in the same segment are equalMeaningIf two angles subtend the same arc from the same side, they have the same measure
The 'segment' is the region between the arc and the chord. All angles inscribed in this region (subtending the same arc) are equal.
Cyclic quadrilateral — opposite angles sum to 180°
A quadrilateral is inscribed in a circle (all four vertices on the circle). What can we say about its opposite angles?
∠A+∠C=180°
TheoremOpposite angles in a cyclic quadrilateral sum to 180°
This is a key property of quadrilaterals inscribed in circles. It means if one angle is 70°, the opposite angle must be 110°.
Worked example — finding unknown angles using circle theorems
A semicircle has diameter AB. Point C lies on the semicircle. If angle CAB = 35°, find angle CBA.
∠ACB=90° (angle in a semicircle)
∠CAB+∠CBA+∠ACB=180° (angles in a triangle)
35°+∠CBA+90°=180°
∠CBA=180° - 90° - 35°=55°
AnswerAngle CBA = 55°
C is on the semicircle and AB is the diameter.
By the angle in a semicircle theorem, angle ACB = 90°.
In triangle ABC, the three angles sum to 180°.
So: angle CAB + angle CBA + angle ACB = 180°.
Substitute: 35° + angle CBA + 90° = 180°.
Solve: angle CBA = 180° - 35° - 90° = 55°.
Try it
Apply circle theorems to find unknown angles. Show which theorem you use in each step.
Question 1
A triangle is inscribed in a semicircle with the diameter as its base. What is the angle at the opposite vertex?
💡 The angle in a semicircle is always 90° (a right angle).
Question 2
Two angles in a circle subtend the same arc. What is their relationship?
💡 Angles in the same segment are equal.
Question 3
A quadrilateral is inscribed in a circle. One angle is 65°. What is the opposite angle?
💡 Opposite angles in a cyclic quadrilateral sum to 180°: 180° - 65° = 115°.
Question 4
In a semicircle with diameter PQ, point R is on the circle. What is angle PRQ?
💡 PQ is the diameter, so angle PRQ (the angle at R) is in a semicircle = 90°.
Question 5
A cyclic quadrilateral has angles 70°, 80° and 120°. Find the fourth angle.
💡 The angle opposite to 90° is 180° - 90° = 90°. Check: 70° + 80° + 120° + 90° = 360°.
Common mistakes
Watch for these when working through the lesson.
Confusing the angle in a semicircle with other angles: only angles inscribed with the diameter as the base are 90°.
Thinking angles in the same segment must be on opposite sides of the chord: they must be on the same side of the chord.
Forgetting that a cyclic quadrilateral's opposite angles sum to 180°: not all angle pairs sum to 180°, only opposite ones.
Related topics
These ideas fit closely with this lesson.
Angles in a circle — chord theorems
Tangent properties
Geometric proof and justification
Practice next
Independent practice will plug in here
This lesson builds the understanding first. Deeper adaptive practice can sit here later.