There are eight circle theorems. They are as follows:

**Circle Theorem 1**: The angle subtended by a chord at the centre is twice the size of the angle subtended by the same chord at the circumference.**Circle Theorem 2**: The angles subtended by a chord in the same segment are equal.**Circle Theorem 3**: The angle of a semi-circle is a right angle.**Circle Theorem 4**: The opposite angles of a cyclic quadrilateral add up to 180^{0}.**Circle Theorem 5**: The angle between a tangent and a radius at the point of contact is a right angle.**Circle Theorem 6**: The two tangents drawn to a circle from a point are equal in length.**Circle Theorem 7**: The angle between a chord and a tangent is equal to the angle subtended by the same chord in the alternate segment.**Circle Theorem 8**: The perpendicular line from the centre to a chord, bisects it.

The following image shows the basics of a circle; they are really important to learn the circle theorems.

You can practise the seven circle theorems with the following simulations; just move the point/s on the circumference and see how they work for yourself. It's fun!

These flash cards will make a significant difference when you revise for your forthcoming exams: very informative and neatly presented; they became best sellers for a reason.

**AS and A Level Only**

The following images show how circle theorems can be proven from basic geometry. Most examination boards in England want students to learn the proofs along with the theorems.

**E.g.1**

If BC = CD and ∠DCF = 27^{0} Find ∠BAC and ∠BCE.

∠CBD = 27^{0} (alternate segment theorem)

∠BDC = 27^{0} (isosceles triangle)

∠BAC = 54^{0} (angle subtended by a chord at the centre = 2 x angle subtended by a chord at the circumference)

∠BAD = 180 -54 = 126^{0} (angle of a triangle add up to 180^{0})

∠BCE = 180 -(126+27) = 27 (angles on a straight line add up to 180^{0})

**E.g.2**

If BA = x, AC = (x - 3), AD = 3cm and EA = 6cm Find x.

∠BEA = ∠ADC (angles subtended in the same segment)

∠EBA = ∠ACD (angles subtended in the same segment)

∠EAB = ∠DAC (vertically opposite angle)

So, EAB and DAC triangles are similar.

BA/AD = EA/AC (corresponding sides are in the same ratio)

x(x - 3) = 6 x 3

x^{2} - 3x - 18 = 0

x = 6 or x = 3.

x = 6cm.

**E.g.3**

Show that BE is parallel to CO.

∠BEC = 180 - (28 + 138) = 14^{0} (angles of a triangle)

∠BOC = 28^{0} (angles subtended by a chord at the centre is twice that of the same at the circumference)

∠CAO = ∠BAE (vertically opposite angle)

∠ACO = 180 - (28 + 138) = 14^{0} (angles of a triangle)

∠BEA = ∠ACO and ∠ABE = ∠AOC (appear to be alternate angles)

So, BE//CO.

**E.g.4**

Find ∠EDC, ∠DFE and ∠ECD.

∠CFB = 180 - 62 = 118^{0} (angles on a straight line)

∠FCB = 180 - (36 + 118) = 26^{0} (angles of a triangle)

∠FDE = ∠FCB (angles subtended by the same chord in the same segment)

∠FDC = 62^{0} (alternate segment theorem)

∠EDC = ∠FDC + ∠FDE = 62 + 26 = 88^{0}

∠DFE = ∠FDE = 26^{0} (FED is an isosceles triangle)

**E.g.5**

Find ∠AOB.

∠ADB + ∠ACB = 180^{0} (opposite angles of a cyclic quadrilateral)

5f + f = 180 => 6f = 180^{0}

f = 30^{0}

∠AOB = 2 x ∠ADB (angles subtended by a chord at the centre is twice that of the same at the circumference)

∠AOB = 60^{0}

**Challenge**

Show that FEGH is a cyclic quadrilateral.

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