Cyclic quadrilaterals · 3 marks

Consider a cyclic quadrilateral. Without drawing its circumcircle, how can we find out whether the centre of the circumcircle lies inside the quadrilateral or outside? What is the best way of finding out?

Answer: Look at the angle each side subtends at an opposite vertex, for example side AB at C (∠ACB). If all four such angles are less than 90°, the centre is inside; if one is 90°, the centre is on that side; if one is more than 90°, the centre is outside, beyond that side. This needs only a protractor, so it is the best way.

Step-by-step solution

Idea: ∠ACB is half the angle that the arc AB (the arc away from C and D) subtends at the centre O. That arc is less than a semicircle exactly when O lies on the same side of AB as the rest of the quadrilateral. A convex quadrilateral contains O exactly when O is on the inner side of all four sides.

For side AB, take the angle it subtends at C: ∠ACB (formed with the diagonal AC). Since C is outside arc AB, ∠ACB = ½ × (angle of arc AB at O).½ mark
If ∠ACB < 90°, arc AB is less than 180°, so O is on the same side of AB as C and D (the inner side). If ∠ACB = 90°, AB is a diameter and O lies on AB. If ∠ACB > 90°, arc AB is more than 180° and O lies on the far side of AB, outside.1 mark
Do this for all four sides: ∠ACB (side AB), ∠BDC (side BC), ∠CAD (side CD), ∠DBA (side DA).½ mark
Rule: all four angles acute ⇒ centre inside; one right angle ⇒ centre on that side; one obtuse ⇒ centre outside. (At most one can be obtuse, since the four arcs total 360°.)1 mark
Best way: this angle test needs only a protractor and the two diagonals. (Another way is to construct the perpendicular bisectors of two sides: they meet at the centre, which can then be seen to be inside or outside; but that takes a construction.)

Measure the angle each side subtends at an opposite vertex (∠ACB, ∠BDC, ∠CAD, ∠DBA). If all are acute, the centre is inside; if one is a right angle, the centre lies on that side; if one is obtuse, the centre is outside, beyond that side. This protractor test is the quickest way.

Answer to write in the exam

Side AB subtends ∠ACB at C; ∠ACB = ½ (angle of arc AB at the centre O).

∠ACB < 90° ⇒ arc AB < 180° ⇒ O on the same side of AB as C, D

∠ACB = 90° ⇒ AB is a diameter ⇒ O on AB

∠ACB > 90° ⇒ arc AB > 180° ⇒ O beyond AB (outside)

Test the four angles ∠ACB, ∠BDC, ∠CAD, ∠DBA.

∴ All acute ⇒ centre inside; one right angle ⇒ centre on that side; one obtuse ⇒ centre outside.

Common mistakes that cost marks

Testing the angles of the quadrilateral itself (∠A, ∠B, …). A rectangle has all angles 90° yet its centre is inside; the test uses the angle between a side and a diagonal.
Thinking the centre is always inside a cyclic quadrilateral. It is outside when one side cuts off more than a semicircle.
Measuring ∠ACB for side BC by mistake; the side must be the one opposite the vertex used.

How this can come in the exam

MCQ (1 mark)

In a cyclic quadrilateral ABCD, side AB subtends ∠ACB = 100° at C. The centre of the circle lies

inside the quadrilateral
on side AB
outside, beyond side AB
at vertex C

Show answer

(C) outside, beyond side AB
∠ACB > 90°, so arc AB is more than a semicircle (200° at the centre) and the centre lies beyond AB, outside the quadrilateral.

Try one yourself

In a cyclic quadrilateral PQRS, ∠PRQ = 40°, ∠QSR = 50°, ∠RPS = 35° and ∠SQP = 55°. Is the centre inside or outside?