A quadrilateral in a circle
A cyclic quadrilateral is a quadrilateral whose four corners all lie on one circle. Name the corners A, B, C and D in order round the circle, so the sides are AB, BC, CD and DA.
Two corners that are not joined by a side are called opposite: A and C are opposite, and so are B and D.
A cyclic quadrilateral ABCD: all four corners are on the circle.
Opposite angles add to 180°
In this quadrilateral the angle at A is 110° and the angle at C is 70°, and 110° + 70° = 180°. The angle at B is 95° and the angle at D is 85°, and 95° + 85° = 180° as well.
This is true of every cyclic quadrilateral: its opposite angles add to 180°. So if one angle is x, the angle opposite it is 180° − x. The four angles of any quadrilateral add to 360°, and here each opposite pair takes exactly half of that.
The angle at A is 110° and the angle at C, opposite it, is 70°: 110° + 70° = 180°.
Why: each angle is half of its own arc
The angle at A is an angle at the circumference. Its arms run to B and D, so it stands on the arc from B to D that passes through C, on the far side from A. The angle at C stands on the other arc from B to D, the one through A.
Join B and D to the center O. By the angle at the center theorem, each angle at the circumference is half of the angle at the center on its own arc. Call the angle at A a and the angle at C c. The angle at the center on A's arc is 2a, and the angle at the center on C's arc is 2c.
Those two angles at the center sit on either side of the radii OB and OD, and together they go all the way round O: 2a + 2c = 360°. Divide by 2: a + c = 180°. The same argument with the radii to A and C shows that the angles at B and D add to 180°.
In the figure, the angle at the center on C's arc is 140° = 2 × 70°, and the reflex angle on A's arc is 360° − 140° = 220° = 2 × 110°.
The angle at A stands on the gold arc from B to D, the arc through C.
The angle at the center between OB and OD, on the side of A, is 140°, twice the 70° at C. The reflex angle on the other side is 220°, twice the 110° at A, and 140° + 220° = 360°.
The exterior angle
Continue the side AB past B, to a point E. The angle CBE, between BC and the continued line, is an exterior angle of the quadrilateral. The angles ABC and CBE make a straight line, so CBE = 180° − ABC. The angle ADC, opposite B, is also 180° − ABC. So the exterior angle at a corner of a cyclic quadrilateral equals the interior angle at the opposite corner.
Here the angle ABC is 95°, so the exterior angle CBE is 180° − 95° = 85°, the same as the angle at D.
Which quadrilaterals are cyclic?
A rectangle is cyclic. Its diagonals are equal and cut each other in half, so the point where they cross is the same distance from all four corners, and the circle with that center passes through all of them. Its opposite angles are 90° + 90° = 180°, as the theorem says.
A parallelogram that is not a rectangle cannot be cyclic. Its opposite angles are equal, and two equal angles that add to 180° would both be 90°. A parallelogram with angles of 70° and 110°, for example, has opposite angles adding to 70° + 70° = 140°, so no circle passes through all four of its corners.
The usual mistakes
Adding angles that are next to each other. It is the opposite angles that add to 180°. The angles at A and B, 110° and 95°, share the side AB and add to 205°.
Copying the angle. Opposite angles are equal only when both are 90°. Otherwise the opposite angle is 180° − x.
Taking 360° − x. 360° is the total of all four angles. One pair of opposite angles adds to 180°.
Worked example: Four Fence Posts Round a Circular Pond
Question Four fence posts A, B, C and D stand in that order on the edge of a circular pond, and straight rails join A to B, B to C, C to D and D to A. Angle DAB is 78° and angle ABC is 105°. (a) Find angle BCD and angle CDA. (b) The rail AB is continued in a straight line past B to a lamp at E. Find angle CBE.
1.All four posts lie on the edge of the pond, a circle, so ABCD is a cyclic quadrilateral and its opposite angles add to 180°.
All four posts stand on one circle, so ABCD is a cyclic quadrilateral and its opposite angles add to 180°. 2.Angle BCD is opposite angle DAB: BCD = 180° − 78° = 102°.
C is opposite A: BCD = 180° − 78° = 102°. 3.Angle CDA is opposite angle ABC: CDA = 180° − 105° = 75°. (a) The angles are 102° at C and 75° at D.
(a) D is opposite B: CDA = 180° − 105° = 75°. 4.ABE is a straight line, so angles ABC and CBE add to 180°: CBE = 180° − 105° = 75°.
ABE is a straight line, so CBE = 180° − 105° = 75°. 5.(b) Angle CBE is 75°, the same as the interior angle at the opposite corner D. Check: the four angles of the quadrilateral add to 78° + 105° + 102° + 75° = 360°.
(b) CBE = 75°, equal to the interior angle at the opposite corner D.
Answer: (a) 102° and 75°; (b) 75°
Common mistakes
- Pairing the angles that are next to each other, 78° and 105°, and expecting them to add to 180°. It is the opposite angles of a cyclic quadrilateral that add to 180°; neighboring angles need not.
- Taking angle CBE to be equal to angle ABC. The two angles lie on a straight line, so they add to 180°; the exterior angle equals the opposite interior angle at D, not the interior angle at B.
More congruence, similarity and circle theorems problems, worked step by step →