# circle theoram diameter circumference radius chord major segment minor segment minor arc major arc...

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• Circle Theoram

• diameterchordMajor SegmentMinor SegmentMinor SectorMajor SectorParts of the CircleParts

• Arc AB subtends angle x at the centre.Arc AB subtends angle y at the circumference.Chord AB also subtends angle x at the centre.Chord AB also subtends angle y at the circumference.Introductory TerminologyTermgy

• Th1

• The angle subtended at the centre of a circle (by an arc or chord) is twice the angle subtended at the circumference by the same arc or chord. (angle at centre)Watch for this one later.

• 42o (Angle at the centre). 70o(Angle at the centre)

• (180 2 x 42) = 96o (Isos triangle/angle sum triangle). 48o (Angle at the centre) 124o (Angle at the centre) (180 124)/2 = 280 (Isos triangle/angle sum triangle).

• 90o angle in a semi-circle90o angle in a semi-circle20o angle sum triangle90o angle in a semi-circle60o angle sum triangleThis is just a special case of Theorem 1 and is referred to as a theorem for convenience.Th2

• Th3

• Angle x = angle y = 38oAngle x = 30oAngle y = 40o

• Th4

• 180 (90 + 36) = 54o Tan/rad and angle sum of triangle.90o angle in a semi-circle60o angle sum triangleIf OT is a radius and AB is a tangent, find the unknown angles, giving reasons for your answers.

• 45o (Alt Seg)60o (Alt Seg)75o angle sum triangleTh5

• The opposite angles of a cyclic quadrilateral are supplementary. (They sum to 180o)Angles y + w = 180oAngles x + z = 180oAngles p + r = 180oAngles q + s = 180oTh6

• 180 85 = 95o (cyclic quad) 180 110 = 70o (cyclic quad) 180 135 = 45o (straight line) 180 70 = 110o (cyclic quad) 180 45 = 135o (cyclic quad)

• From any point outside a circle only two tangents can be drawn and they are equal in length.Th7

• 90o (tan/rad)PTQOxowo98oyozoPQ and PT are tangents to a circle with centre O. Find the unknown angles giving reasons.90o (tan/rad)49o (angle at centre)360o 278 = 82o (quadrilateral)

• 90o (tan/rad)PTQOyo50oxo80oPQ and PT are tangents to a circle with centre O. Find the unknown angles giving reasons.180 140 = 40o (angles sum tri)50o (isos triangle) 50o (alt seg)wozo

• OS = 5 cm (pythag triple: 3,4,5)Th8

• Angle SOT = 22o (symmetry/congruenncy)Angle x = 180 112 = 68o (angle sum triangle)

• 65o (Alt seg)130o (angle at centre)25o (tan rad)25o (isos triangle)Mixed Q 1

• 22o (cyclic quad)68o (tan rad)44o (isos triangle)68o (alt seg)Mixed Q 2

• Extend AO to DAO = BO = CO (radii of same circle) Triangle AOB is isosceles(base angles equal) Triangle AOC is isosceles(base angles equal) Angle AOB = 180 - 2 (angle sum triangle) Angle AOC = 180 - 2 (angle sum triangle) Angle COB = 360 (AOB + AOC)(
• To prove that angle CAB = angle BDCWith centre of circle O draw lines OB and OC.Angle COB = 2 x angle CAB (Theorem 1).Angle COB = 2 x angle BDC (Theorem 1).2 x angle CAB = 2 x angle BDCAngle CAB = angle BDCQEDProof 3

• Proof 4

• To prove that OT is perpendicular to ABAssume that OT is not perpendicular to ABThen there must be a point, D say, on AB such that OD is perpendicular to AB.Since ODT is a right angle then angle OTD is acute (angle sum of a triangle).But the greater angle is opposite the greater side therefore OT is greater than OD.But OT = OC (radii of the same circle) therefore OC is also greater than OD, the part greater than the whole which is impossible.Therefore OD is not perpendicular to AB.By a similar argument neither is any other straight line except OT.Therefore OT is perpendicular to AB.QEDTo prove that the angle between a tangent and a radius drawn to the point of contact is a right angle.OABTTheorem 4

• To prove that angle BTD = angle TCDWith centre of circle O, draw straight lines OD and OT.Let angle DTB be denoted by .Then angle DTO = 90 - (Theorem 4 tan/rad)Also angle TDO = 90 - (Isos triangle)Therefore angle TOD = 180 (90 - + 90 - ) = 2 (angle sum triangle) 2Angle TCD = (Theorem 1 angle at the centre) Angle BTD = angle TCD QEDProof 5

• To prove that angles A + C and B + D = 1800Draw straight lines AC and BDChord DC subtends equal angles (same segment) Chord AD subtends equal angles (same segment) Chord AB subtends equal angles (same segment) Chord BC subtends equal angles (same segment) 2( + + + ) = 360o (Angle sum quadrilateral) + + + = 180o Angles A + C and B + D = 1800QEDProof 6

• To prove that AP = BP.With centre of circle at O, draw straight lines OA and OB.OA = OB (radii of the same circle)Angle PAO = PBO = 90o (tangent radius).Draw straight line OP.In triangles OBP and OAP, OA = OB and OP is common to both.Triangles OBP and OAP are congruent (RHS)Therefore AP = BP.QEDProof 7

• To prove that AB = BC.From centre O draw straight lines OA and OC.In triangles OAB and OCB, OC = OA (radii of same circle) and OB is common to both.Angle OBA = angle OBC (angles on straight line)Triangles OAB and OCB are congruent (RHS)Therefore AB = BCQEDProof 8

• Worksheet 1Parts of the Circle

• Worksheet 2Th1Measure the angle subtended at the centre (y) and the angle subtended at the circumference (x) in each case and make a conjecture about their relationship.

• To Prove that the angle subtended by an arc or chord at the centre of a circle is twice the angle subtended at the circumference by the same arc or chord.OCBATheorem 1 and 2Worksheet 3

• To Prove that angles subtended by an arc or chord in the same segment are equal.ATheorem 3Worksheet 4

• To prove that the angle between a tangent and a radius drawn to the point of contact is a right angle.OABTTheorem 4Worksheet 5

• To prove that the angle between a tangent and a chord through the point of contact is equal to the angle subtended by the chord in the alternate segment.Theorem 5ATBCDOWorksheet 6

• ABDCTo prove that the opposite angles of a cyclic quadrilateral are supplementary (Sum to 180o).Theorem 6AlphaBetaChideltaWorksheet 7

• Worksheet 8To prove that the two tangents drawn from a point outside a circle are of equal length.Theorem 7OABP

• To prove that a line, drawn perpendicular to a chord and passing through the centre of a circle, bisects the chord. Theorem 8OABCWorksheet 9

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