CIRCLE THEOREMS

Questions normally ask for the size of an angle and reasons

why.

In each case use the theorems you know and what is special.

Any triangles made from radii will always be isosceles.

CIRCLE THEOREMS

Tangent and Radius are perpendicular

CIRCLE THEOREMS

Angle in a Semi-Circle is always 90°

CIRCLE THEOREMSThe angle at the Centre is twice

that at the circumference

CIRCLE THEOREMS

Angles from the same chord are equal

Angles A and C are equal from the chord BD

In the same way angle B and D are equal, coming from the chord AC

CIRCLE THEOREMSThe opposite angles in a cyclic

quadrilateral add to 180°

A and C add to 180 and so do B and D

CIRCLE THEOREMSAlternate Segment

theoremAngle between the tangent and the chord (angle a on the diagram) equals the angle in the alternate(other) segment – angle b

Examples

Try the following examples. The question is on one page, go to next

page for a solution

CIRCLE THEOREMSExamples

CIRCLE THEOREMSExamples

In Δ OAB angles A and B are equalOA=OB as radii which makes isosceles

Angle OBA will then be (180-2x) ÷ 2 which is (90-x)

OB and OT are perpendicular as they are radius and tangent and so Angle ABT will be 90-(90-x) which is x

CIRCLE THEOREMSExamples

CIRCLE THEOREMSExamples

Draw the line AO and label angles at B and C as b and c

In Δ OAC, OA=OC as radii so Δ is isosceles\ Angle OAC = OCA = c\ Angle AOC is 180 – 2c

In same way in ΔOBC angle BOC is 180 – 2b

Angles around a point are 360 so angle BOC=360- AOC – BOC= 360 – (180 -2b) –(180-2c)= 2b + 2c= 2 x ( b + c) = 2 x angle CAB

cb

CIRCLE THEOREMSExamples

The diagram shows a circle centre O. A, B and C are points on the circumference. DCO is a straight line.DA is a tangent to the circle. Angle ADO = 36°

(a) Work out the size of angle AOD°(b) Work out the size of angle ABC.

Give reasons for your answers

CIRCLE THEOREMSa. Angle OAD is 90 as OA

is a radius and AD a tangent

b. Angle AOD is 54° as OAD is a triangle and OAD is 90.

Angle ABC is the angle at the circumference when AOC is at the centre and so is 27°

Angle at centre theorem

CIRCLE THEOREMSExamples

CIRCLE THEOREMSAngle PQT = angle PTA = 58°Alternate segment theorem

In Δ PQT, PQ=QT and so it isosceles and the angles at P and T are equal and as PQT is 58, they are (180-58) ÷ 2 which is 61

Angle POT is 90-58 = 32 because angle OAT is 90 (radius/tangent)

Finally OTQ = 61-32 = 29°

CIRCLE THEOREMSExamples

CIRCLE THEOREMSAD is a chord and angles ABD and ACD are from that chord and so are equal. ACD = 54°

CIRCLE THEOREMSExamples

CIRCLE THEOREMSABCD is a cyclic quadrilateral and so angle BAD = 50

Angle BOD is the angle at the centre and so is 100°

In the quadrilateral OBDC we now have 100 and 130 which leaves 360-230 = 130 for the other two.

As OB=OD (radii), OBDC is symmetric and angle OBC and ODC are equal and must be 130 ÷ 2 = 65°

CIRCLE THEOREMSExamples

CIRCLE THEOREMSABCD is a cyclic quadrilateral and so angle D is 52°

Angle AOC is the angle at the centre and so will be twice that at the circumference and so is 2 x 52 = 104°

CIRCLE THEOREMSExamples

CIRCLE THEOREMSAs no angles are known set the one you want to x

BC=BD so Δ BCD is isosceles and so BCD is also x

Angle C is 90 as it is the angle in a semi-circle and so angle ACD is 90 - x

Because AC and BD are parallel, angles ACD and CDB are equal (alternate angles) and so x = 90 – x

Solving gives x = 45°