Angles and Circles 2

Angles, Circles and Tangents

There are a number of important geometric results based on angles in circles. (The first you have met already.)

Any angle subtended at the circumference from a diameter is a right angle.

The angle subtended by an arc, PQ, at the centre is twice the angle subtended at the circumference.

Proof

OP = OC (equal radii), so

angle CPO = angle PCO (= x, say).

Similarly,

angle CQO = angle QCO (= y, say).

Now, extending the line CO to D, say, note that

angle POD		= x + x
	= 2x

and, similarly,

angle QOD	= y + y
	= 2y

Hence,

angle POQ	= 2x + 2y
	= 2 (x + y)
	= 2 × angle PCQ

as required.

Angles subtended at the circumference by a chord (on the same side of the chord) are equal; that is, in the diagram a = b.

Proof

The angle at the centre is 2a or 2b (according to the first result).

Thus 2a = 2b or a = b, as required.

In cyclic quadrilaterals (quadrilaterals where all 4 vertices lie on a circle), opposite angles sum to 180° ; that is

a + c = 180°

and

b + d = 180°

Proof

Construct the diagonals AC and BD, as below.

Then label the angles subtended by AB as w; that is

angle ADB = angle ACB (= w)

Similarly for the other chords, the angles being marked x, y and z as shown.

Now, in triangle ABD, the sum of the angles is 180° ,
so

w + z + (x + y) = 180°

You can rearrange this as

(x + w) + (y + z) = 180°

which shows that

angle CDA + angle CBA = 180°

proving one of the results.

The other result follows in a similar way.

Worked Examples

Find the angles marked in the diagrams. In each case O is the centre of the circle.

(a)

Show me

As both angles are drawn on the same chord, the angles are equal, so

a = 35°

(b)

Show me

Angle b and the 25°angle are drawn on the same chord, so

b = 25°

Angle a is drawn at the centre O on the same chord as the 25°angle, so

a	= 2 × 25°
	= 50°

Find the angles marked in the diagrams. O is the centre of the circle.

(a)

Show me Opposite angles in a cyclic quadrilateral add up to 180° . So

a + 80°	= 180°
a	= 100°

and

b + 110°	= 180°
b	= 70°

(b)

Show me Consider the angles c and 210° . Since the angle at the centre is double the angle in a segment drawn on the same arc,

2c	= 210°
c	= 105°

Angles c and e add up to 180° because they are opposite angles in a cyclic quadrilateral.

c + e	= 180°
105°+ e	= 180°
e	= 180° − 105°
	= 75°

Consider the quadrilateral BODC. The four angles in any quadrilateral add up to 360°. So

d + e + 210°+ 20°	= 360°
d	= 360° − 210° − 20° − e
	= 130° − c
	= 130° − 75°
	= 55°

In the diagram the line AB is a diameter and O is the centre of the circle. Find the angles marked.

Show me

Consider triangle OAC. Since OA and OC are radii, triangle OAC is isosceles. So

a = 50°

The angles in a triangle add to 180° , so for triangle OAC,

a + b + 50°	= 180°
b	= 180° − 50° − a
	= 80°

Since AB is a diameter of the circle, the angle ACB is a right angle, so

a + 20° + c	= 90°
c	= 90° − 20° − a
	= 20°

Angle d and angle OAC are angles in the same segment, so

d	= angle OAC
	= 50°

Angle e is drawn on the same arc as the angle at the centre, AOC, so

b	= 2e
e	= b
	= 40°

In the diagram the chords AB and CD are parallel. Prove that the triangles ABE and DEC are isosceles.

Show me

Angles a and BDC are angles in the same segment, so

angle BDC = a

Since AB and DC are parallel, angles a and ACD are equal alternate angles,

angle ACD = a = angle BDC

Hence in triangle DEC, the base angles at C and D are equal, so the triangle is isosceles.

The angle at B, angle ABD, equals the angle at C, angle ACD, because they are angles in the same segment:

angle ABD = angle ACD = a

Hence triangle ABE is isosceles, since the angles at A and B are equal.

Proof

Proof

Proof

Worked Examples

Exercises