If a line segment AB joining two points A, B subtends equal angles at two other points C, D that lie on the same side of AB, then the four points lie on a circle.
Step-by-step solution
To find: Show that A, B, C, D lie on one circle
Idea: There is exactly one circle through the three non-collinear points A, B, C. Show D cannot be outside it or inside it, using two facts: angles in the same segment are equal, and an exterior angle of a triangle is bigger than each opposite interior angle.
- A, B, C are not collinear, so there is a circle through A, B and C. Suppose D is not on it; then D is either outside or inside the circle.½ mark
- D outside: AD cuts the circle at a point E between A and D. C and E are on the same arc cut off by AB, so ∠AEB = ∠ACB (angles in the same segment).1 mark
- ∠AEB is an exterior angle of ΔBED, so ∠AEB > ∠EDB = ∠ADB. Then ∠ACB = ∠AEB > ∠ADB = ∠ACB, so ∠ACB > ∠ACB: impossible.1 mark
- D inside: extend AD to meet the circle at E. Again ∠AEB = ∠ACB (same segment).1 mark
- Now ∠ADB is an exterior angle of ΔBDE, so ∠ADB > ∠DEB = ∠AEB = ∠ACB. But ∠ADB = ∠ACB (given): impossible again.1 mark
- Both options are ruled out, so D lies on the circle through A, B, C. A, B, C, D are concyclic.½ mark
Answer to write in the exam
Given: C, D on the same side of AB, ∠ACB = ∠ADB. To prove: A, B, C, D are concyclic.
Draw the circle through A, B, C (A, B, C non-collinear). Suppose D is not on it.
Case 1, D outside: AD meets the circle at E.
∠AEB = ∠ACB (angles in the same segment)
∠AEB > ∠ADB (exterior angle of ΔBED)
⇒ ∠ACB > ∠ADB = ∠ACB, a contradiction.
Case 2, D inside: AD produced meets the circle at E.
∠ADB > ∠AEB (exterior angle of ΔBDE) = ∠ACB (same segment)
⇒ ∠ADB > ∠ACB = ∠ADB, a contradiction.
∴ D lies on the circle through A, B, C; A, B, C, D are concyclic.
Common mistakes that cost marks
- Leaving out the condition “on the same side of AB”. Points on opposite sides seeing AB at equal angles are not concyclic in general.
- Treating only one case (outside) and saying “similarly” without working the inside case, where the exterior angle is at D instead of E.
- Getting the exterior angle inequality backwards: an exterior angle is greater than each interior opposite angle.
How this can come in the exam
Assertion (A): If ∠APB = ∠AQB = 50° and P, Q lie on the same side of AB, then A, B, P, Q are concyclic.
Reason (R): A line segment that subtends equal angles at two points on the same side of it has its end points and those two points on one circle.
- Both A and R are true, and R is the correct explanation of A.
- Both A and R are true, but R is not the correct explanation of A.
- A is true, but R is false.
- A is false, but R is true.
Show answer
(A) Both A and R are true, and R is the correct explanation of A.
R is the concyclicity result, and A is a direct use of it.
In quadrilateral ABCD, the diagonals meet at O, ∠BAC = 35° and ∠BDC = 35°. Show that A, B, C, D are concyclic.
Show answer
Segment BC subtends ∠BAC = 35° at A and ∠BDC = 35° at D (1 mark). A and D lie on the same side of BC (both are vertices of the quadrilateral opposite side BC), so A, B, C, D lie on a circle (1 mark).Try one yourself
P and Q are on the same side of a segment XY, with ∠XPY = 72° and ∠XQY = 72°. Are X, Y, P, Q concyclic? What if ∠XQY were 70°?
Show answer
With 72° and 72°: yes, equal angles on the same side make them concyclic. With 70°: no; the angles differ, so Q is not on the circle through X, Y, P (it lies outside it).
More questions like this
- The sum of two opposite angles of a cyclic quadrilateral is 180°.
- A cyclic quadrilateral has angles measuring ∠A = 80°, ∠B = 110°, ∠C = 100°, and ∠D = 70°. Can such a quadrilateral be drawn? Explain why or why not.
- If two opposite angles of a quadrilateral add up to 180°, then the vertices of the quadrilateral lie on a circle, i.e., they are concyclic.
- In a circle, a chord is 5 cm away from the centre. If the radius of the circle is 13 cm, what is the length of the chord?
- An arc of a circle subtends an angle of 70° at the centre. What is the measure of the angle subtended by the arc at a point on the circle?