Combining Circle Theorems

Worked example

1. 1

   $\angle PQR=90^{\circ }$ — PR is a diameter and Q is on the circle, so angle in a semicircle is 90°.

2. 2

   $\angle PSQ=\angle PRQ=31^{\circ }$ — S and R both stand on chord PQ, on the same side of it, so angles in the same segment are equal.

Worked example

1. 1

   $\angle BAD=180^{\circ }-124^{\circ }=56^{\circ }$ — A and C are opposite corners of ABCD, so opposite angles of a cyclic quadrilateral add up to 180°.

2. 2

   $\angle BOD=2\times 56^{\circ }=112^{\circ }$ — O and A both look at chord BD, so angle at the centre is twice the angle at the circumference.

3. 3

   $\angle OBD=\angle ODB$ — OB and OD are radii, so triangle OBD is isosceles: base angles of an isosceles triangle are equal.

4. 4

   $\angle OBD+\angle ODB=180^{\circ }-112^{\circ }=68^{\circ }$ — Angles in a triangle add up to 180°, so the two base angles share what is left.

5. 5

   $\angle OBD=68^{\circ }÷2=34^{\circ }$ — The two base angles are equal, so each one is half of 68°.

Worked example

1. 1

   $\angle ACB=90^{\circ }$ — AB is a diameter and C is on the circle, so angle in a semicircle is 90°.

2. 2

   $\angle ABC=180^{\circ }-90^{\circ }-27^{\circ }=63^{\circ }$ — Triangle ABC now has two known angles, and angles in a triangle add up to 180°.

3. 3

   $\angle ADC=\angle ABC=63^{\circ }$ — D and B both stand on chord AC, on the same side of it, so angles in the same segment are equal.

Worked example

1. 1

   $\angle BED=\angle ABE=68^{\circ }$ — AB is parallel to ED, and BE crosses both in a Z shape, so alternate angles are equal.

2. 2

   $\angle BDE=68^{\circ }$ — The 68° is between tangent AB and chord BE. D is across chord BE, so by the alternate segment theorem it gets the same angle.

3. 3

   $\angle BED+\angle BDE=68^{\circ }+68^{\circ }=136^{\circ }$ — Add the two angles you know in triangle BDE.

4. 4

   $\angle EBD=180^{\circ }-136^{\circ }=44^{\circ }$ — Angles in a triangle add up to 180°, so the third angle is what is left.

5. 5

   $\angle DCE=\angle EBD=44^{\circ }$ — C and B both stand on chord DE, on the same side of it, so angles in the same segment are equal.

Worked example

1. 1

   $\angle ACB=64^{\circ }$ — The 64° is between tangent AT and chord AB. C is across chord AB, so by the alternate segment theorem it gets the same angle.

2. 2

   $\angle ABC=\angle ACB=64^{\circ }$ — AB = AC, so triangle ABC is isosceles and base angles of an isosceles triangle are equal.

3. 3

   $\angle BAC=180^{\circ }-64^{\circ }-64^{\circ }=52^{\circ }$ — Angles in a triangle add up to 180°, so the angle at A is what is left.

4. 4

   $\angle BDC=180^{\circ }-52^{\circ }=128^{\circ }$ — A and D are opposite corners of ABDC, so opposite angles of a cyclic quadrilateral add up to 180°.

Worked example

1. 1

   $\angle OAD=\angle OCD=90^{\circ }$ — OA and OC are radii that meet the tangents, so tangent is perpendicular to the radius.

2. 2

   $\angle AOC=360^{\circ }-90^{\circ }-90^{\circ }-50^{\circ }=130^{\circ }$ — OADC has four angles, and angles in a quadrilateral add up to 360°.

3. 3

   $\text{reflex }\angle AOC=360^{\circ }-130^{\circ }=230^{\circ }$ — The reflex angle is the rest of the turn round O: angles at a point add up to 360°.

4. 4

   $\angle ABC=130^{\circ }÷2=65^{\circ }$ — B is on the big arc, so it looks at the same arc as the 130°. Angle at the centre is twice the angle at the circumference, so halve it.

5. 5

   $\angle AEC=230^{\circ }÷2=115^{\circ }$ — E is on the small arc, so it looks at the other side, the reflex 230°. Halve that.

Worked example

1. 1

   $\angle OBA=x$ — OA and OB are radii, so triangle OAB is isosceles: base angles of an isosceles triangle are equal.

2. 2

   $\angle AOB=180-x-x=180-2x$ — Angles in a triangle add up to 180°. Collect the two x's: x + x = 2x.

3. 3

   $\angle ACB=\frac{180-2x}{2}$ — C looks at the same chord AB as O. Angle at the centre is twice the angle at the circumference, so halve it.

4. 4

   $\angle ACB=90-x$ — Halve each part: 180 ÷ 2 = 90 and 2x ÷ 2 = x.

5. 5

   $90-x=x+20$ — Both expressions are angle ACB, so set them equal.

6. 6

   $70=2x$ — Add x to both sides and take 20 from both sides.

7. 7

   $x=35$ — Divide by 2. Check: OAB = 35°, AOB = 110°, ACB = 55° = 35° + 20°.

Worked example

1. 1

   $\angle EBF=\angle EDF=52^{\circ }$ — B and D both stand on chord EF, on the same side of it, so angles in the same segment are equal.

2. 2

   $\angle BEF=\angle FBC=64^{\circ }$ — The 64° is between tangent BC and chord BF. E is across chord BF, so by the alternate segment theorem it gets the same angle.

3. 3

   $\angle BFE=180^{\circ }-52^{\circ }-64^{\circ }=64^{\circ }$ — Triangle BEF now has two known angles, and angles in a triangle add up to 180°.

4. 4

   $\angle BFE=\angle FBC=64^{\circ }$ — These two angles sit in a Z shape across BF and they are equal. So they are alternate angles, and ABC is parallel to EF.