Two circle theorems do more work in GCSE geometry papers than almost any others: the angle in a semicircle is a right angle, and the opposite angles of a cyclic quadrilateral add to 180°. Learn them properly and a whole family of questions, some of which look quite intimidating at first glance, collapses into three or four lines of arithmetic. The trick is not memorising the wording. It is learning to see the shape, name the reason, and write it down in the form examiners reward.
Start with the two shapes that matter
Nearly every question in this topic begins with one of two pictures. Either a triangle sits inside a circle with one side passing through the centre, or four points on a circle are joined up to make a quadrilateral. Spot which one you have and you already know the form of the answer.
- Semicircle triangle: a diameter, plus any point on the circumference, joined to both ends.
- Cyclic quadrilateral: four points on the circumference, joined in order, with every vertex sitting on the circle.
The word cyclic is the clue. It means a single circle passes through all four vertices. Push one corner inside the circle and the shape is no longer cyclic, so the 180° rule does not apply to it.
Theorem 1: the angle in a semicircle is 90°
Draw a circle with centre O. Mark the two ends of a diameter as A and B, then choose any point C on the circumference and join A to C and B to C. Wherever you put C, the angle at C is exactly 90°.
The reason to write is: the angle in a semicircle is a right angle. Many mark schemes also accept “the angle subtended by a diameter at the circumference is 90°”.
Why it works
The proof needs only radii and isosceles triangles. Draw the radius from O to C. Now OA, OB and OC are all radii, so they are equal in length.
Triangle AOC is isosceles because OA = OC, so angle OAC = angle ACO. Call both x. Triangle BOC is also isosceles because OB = OC, so angle OBC = angle OCB. Call both y. The three angles of triangle ABC are x, (x + y) and y, and they sum to 180°. So 2x + 2y = 180, giving x + y = 90 — exactly the angle at C.
Notice what the proof does not use: no trigonometry, no coordinates, no measurements. Circle theorem proofs are almost always built from radii and isosceles triangles, which is why drawing the radius to the key point is such a reliable first move.
Spotting it under pressure
Look for a side of a triangle that passes through the centre of the circle. If it does, and the third vertex sits on the circumference, you have been handed a free right angle. Add a small square in the corner, write your reason, and carry on. Questions that then ask for a remaining angle become simple subtraction from 180°.
Theorem 2: opposite angles of a cyclic quadrilateral
Take four points A, B, C and D on a circle and join them in order. The shape ABCD is a cyclic quadrilateral. Angles A and C are opposite each other, as are B and D, and each pair adds to 180°.
Write it as: opposite angles of a cyclic quadrilateral sum to 180°. If a question extends one side instead, the exterior angle equals the interior opposite angle. That is the same fact wearing a different hat.
A proof you can sketch quickly
Each opposite angle is also an angle at the circumference. The chord closing the quadrilateral subtends one angle at the centre for one pair of opposite angles, and the reflex angle at the centre for the other pair. Those two angles at the centre complete a full turn, so together they make 360°. Using the angle-at-the-centre theorem, twice angle A plus twice angle C equals 360°, and halving gives A + C = 180°. The argument for B and D is identical.
Two GCSE-style examples
Example 1: a diameter and a triangle
A, B and C lie on a circle with AB as a diameter. Angle BAC = 32°. Find angle ABC.
Angle ACB = 90°, because the angle in a semicircle is a right angle. The angles in triangle ABC sum to 180°, so angle ABC = 180 − 90 − 32 = 58°.
Example 2: a cyclic quadrilateral with algebra
ABCD is a cyclic quadrilateral. Angle A = 3x + 10 and angle C = x + 30. Find x and the size of angle A.
Opposite angles sum to 180°, so (3x + 10) + (x + 30) = 180. That gives 4x + 40 = 180, so 4x = 140 and x = 35. Angle A = 3 × 35 + 10 = 115°. Check by finding angle C = 35 + 30 = 65°, and 115 + 65 = 180. It works.
Five mistakes that cost easy marks
- Quoting the theorem without naming it. A correct number with no reason can lose marks. Write the full statement, not just “semicircle”.
- Assuming every quadrilateral is cyclic. Check that the vertices genuinely lie on the circle.
- Pairing the wrong angles. In ABCD, A pairs with C and B pairs with D. Opposite means diagonally across, not next door.
- Confusing the exterior angle rule. The exterior angle of a cyclic quadrilateral equals the interior opposite angle, not 180° minus it.
- Rounding too early. Keep exact values until the final line, particularly when algebra is involved.
Make the diagrams work for you
Circle theorem marks are lost on diagrams as often as on calculations. Before you write anything, redraw the figure larger than the printed version and mark the centre clearly with O. Label every vertex you mention in your reasoning, add the right-angle square, and tick the angles you are comparing. If a diagonal or radius is missing but would connect the given information, draw it in — very often that single line is the whole solution. Finish with the layout examiners expect: working, answer, reason.
A good habit is to cover the labels on a diagram and name the theorem you would need before doing any arithmetic. Ten minutes of that beats an hour of copying out proofs, because recognising the shape is the part that has to be instant.
Photo: Carlos Yanez / Pexels