Let A and B be two points on a circle with centre O.
- (i) Are there points X, Y on the circle, on the same side of AB, such that ∠AXB is different from ∠AYB?
- (ii) Is it true that if ∠AXB = ∠AYB, then X and Y lie on the same side of the circle?
- (iii) If ∠AXB = ∠AYB, and X and Y do not lie on the circle, does the circle through A, B and X also pass through Y?
Step-by-step solution
Idea: A point X on the circle sees AB at half the angle that the arc on the other side of AB subtends at the centre. The two arcs cut off by AB subtend angles at O that add up to 360°, so points on the two sides see AB at angles adding up to 180°.
(i) Are there points X, Y on the circle, on the same side of AB, such that ∠AXB is different from ∠AYB?
- X and Y on the same side of AB are both outside the arc AB on the other side. Each of ∠AXB and ∠AYB is half the angle that this arc subtends at O. So ∠AXB = ∠AYB always: no such points exist.1 mark
(ii) Is it true that if ∠AXB = ∠AYB, then X and Y lie on the same side of the circle?
- Read “side” as “side of AB” (X and Y are on the circle). If X and Z are on opposite sides of AB, then ∠AXB = ½(angle of one arc at O) and ∠AZB = ½(angle of the other arc at O). The two arcs together subtend 360°, so ∠AXB + ∠AZB = 180°.½ mark
- So equal angles on opposite sides are possible only when each is 90°, that is, when AB is a diameter. Hence the statement is true when AB is not a diameter, and false when AB is a diameter (every point of the circle then sees AB at 90°).½ mark
(iii) If ∠AXB = ∠AYB, and X and Y do not lie on the circle, does the circle through A, B and X also pass through Y?
- If X and Y are on the same side of AB: AB subtends equal angles at X and Y, so A, B, X, Y lie on one circle (if a segment subtends equal angles at two points on the same side of it, the four points are concyclic). So yes, the circle through A, B, X passes through Y.½ mark
- If X and Y are on opposite sides of AB: points of the circle through A, B, X on Y’s side see AB at 180° − ∠AXB. So Y is on it only if ∠AXB = 90°; in general, no.½ mark
Answer to write in the exam
(i)
X, Y on the circle, same side of AB ⇒ both lie outside the arc AB on the other side
∠AXB = ½(angle subtended by that arc at O) = ∠AYB (angle at the centre is double the angle at the circle)
∴ No; ∠AXB is always equal to ∠AYB.
(ii)
X, Z on opposite sides of AB: ∠AXB + ∠AZB = ½ × 360° = 180°
If ∠AXB = ∠AZB, then each = 90° ⇒ AB is a diameter
∴ True if AB is not a diameter; false if AB is a diameter (all angles 90°).
(iii)
Same side of AB: ∠AXB = ∠AYB ⇒ A, B, X, Y concyclic (equal angles on the same side of AB)
∴ Yes, the circle through A, B, X passes through Y.
Opposite sides: the circle through A, B, X needs ∠AYB = 180° − ∠AXB ⇒ only if both are 90°
∴ In general, no.
Common mistakes that cost marks
- In (i), answering “yes, X nearer to A gives a bigger angle”. Angles in the same segment are all equal, wherever the point is on that arc.
- In (ii), forgetting the diameter case, where every point gives 90°.
- In (iii), ignoring the condition “on the same side of AB”, which the concyclicity result needs.
How this can come in the exam
A and B are points on a circle. X and Y are points of the circle on the same side of AB with ∠AXB = 38°. Then ∠AYB is
- 19°
- 38°
- 76°
- 142°
Show answer
(B) 38°
Angles in the same segment are equal, so ∠AYB = 38°.
Try one yourself
A, B, X, Z lie on a circle, with X and Z on opposite sides of AB. If ∠AXB = 64°, find ∠AZB.
Show answer
Points on opposite sides of AB see it at angles adding to 180°: ∠AZB = 180° − 64° = 116°.
More questions like this
- Find x in the figure.
- 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.
- 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.