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## CIRCLE THEOREMS

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**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 THEOREMS**The angle at the Centre is twice that at the circumference**CIRCLE THEOREMS**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 Angles from the same chord are equal**CIRCLE THEOREMS**A and C add to 180 and so do B and D The opposite angles in a cyclic quadrilateral add to 180°**CIRCLE THEOREMS**Angle between the tangent and the chord (angle a on the diagram) equals the angle in the alternate(other) segment – angle b Alternate Segment theorem**Examples**Try the following examples. The question is on one page, go to next page for a solution**CIRCLE THEOREMS**Examples**In Δ OAB angles A and B are equal**OA=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 THEOREMS Examples**CIRCLE THEOREMS**Examples**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 CIRCLE THEOREMS b c Examples**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° CIRCLE THEOREMS Examples (a) Work out the size of angle AOD° (b) Work out the size of angle ABC. Give reasons for your answers**Angle OAD is 90 as OA is a radius and AD a tangent**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 THEOREMS**CIRCLE THEOREMS**Examples**Angle 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 THEOREMS**CIRCLE THEOREMS**Examples**AD is a chord and angles ABD and ACD are from that chord and**so are equal. ACD = 54° CIRCLE THEOREMS**CIRCLE THEOREMS**Examples**ABCD 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 THEOREMS**CIRCLE THEOREMS**Examples**ABCD 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 THEOREMS**CIRCLE THEOREMS**Examples**As 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° CIRCLE THEOREMS