Chapter 3. - parts of a circle.
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1 Chapter 3 - parts of a circle. 3.1 properties of circles. - area of a sector of a circle. the area of the smaller sector can be found by the following formula: A = qº 360º pr2, given q in degrees, or! A = q 2p pr2, given q in radians - the Chord Right Bisector Property (RBP). OM^AB iff AM = MB ie. the centre of a circle always lies on the perpendicular bisector of any chord. I. given OM^AB because DOAM DOBM (HS ), therefore, AM = MB ( Ds)
2 II. given AM = MB because DOAM DOBM (SSS), q = j ( Ds) therefore, q = j = 90º ex. prove that XY is the right bisector of PQ. DXPY DXQY (SSS) therefore PXM = QXM = q so DXPM DXQM (SAS ) ergo, XMP = XMQ = 90º and PM = MQ ( Ds) QED - the Equal Chords Property (ECP). AB = CD iff d 1 = d 2 I. given AB = CD ND = MB DOBM DODN OM = ON ( Ds) II. given OM = ON DOMB DONC (HS ) therefore, MB = NC similarly, AM = DN AM + MB = DN + NC thus AB = DC
3 3.2 angles in a circle. - some definitions. q is called the central angle, and it is said to be subtended by minor arc AB. the reflex angle AOB is the central angle subtended by major arc AB. AXB is an inscribed angle subtended by arc AYB. AYB is an inscribed angle subtended by arc AXB. - Angles in a Circle Properties (ACP). 1. central angles are twice the size of inscribed angles subtended by the same arc. 2. inscribed angles subtended by the same arc are equal. 3. inscribed angles subtended by a diameter equal 90º. because OC = OA = OB, DOAC and DOCB are isosceles thus, ACO = CAO and OBC = OCB by EAT, DOA = 2 OAC AOB = 2 OAC ergo, = 2 ACB ( ) and DOB = 2 ( OCB) ( ) + 2 ( OCB) QED. ( )
4 ex. find the value of w, x, y, z. answers: w = 40º, x = 70º, y = 60º, z = 30º ex. prove DACE is isosceles. in DEBD, ED = BD ex. prove that AB = BC. therefore d = e (ITT) but g = d, ergo g = d = e a = e because a and e are subtended by the same arc therefore a = g by transitivity, hence DACE is isosceles. QED. because OA and OC are radii of a common circle, OA = OC. OB is common to both DOBC and DOBA. because OA is a diameter, by ACP, OBA = 90º because OB^AC, \ AB = BC (RBP). ex. determine the size of the central angle subtended by an arc of 2p if the circle s radius is 6. q 360º = 2p 12p C = pd =12p, therefore q 360º = 1 6 q = 60º 3.3 cyclic quadrilaterals. - concyclic points are points which lie on the circumference of the same circle. - cyclic polygons have concyclic vertices. ex. prove that all triangles are cyclic given any DABC ie. prove OA = OB = OC clearly, OA = OB (RBT) similarly, OB = OC (RBT) by transitivity, therefore OA = OB = OC QED.
5 ex. prove that if a quadrilateral ABCD is cyclic, then its opposite angles are supplementary. 2q + 2b = 360º q + b =180º QED. - Cyclic Quadrilateral Properties. 1. opposite angles are supplementary 2. any exterior angle is equal to the opposite interior angle 3. a side subtends equal angles at the remaining vertices ie. BAC = BDC or CAD = CBD ex. in a cyclic quadrilateral ABCD, AB = AD, BCD =110º, BAC = 30º. find ABC. a = 40º (CQP) and j = 55º but CD subtends both CAD and CBD so CAD = CBD = 40º therefore, ABC = 55º+40º+95º ex. show that BCED is cyclic. in DABE, A = 70º in DABC, 50º+70º+2q =180º q = 30º BEC =100º= BDC because side BC subtends equal angles at D and E, therefore BCED is cyclic.
6 3.5 tangent properties. - Tangent/Radius Properties (TRP) AB is a tangent iff OP^AB I. given AB is a tangent assume OPA 90º WLOG let OPA < 90º construct Q on AB such that OQP = OPA ergo, OQ = OP (ITT) Æ therefore OP^AB II. given OP^BA assume AB is a secant. clearly, OQ = OP so OQP = OPQ = 90º Æ therefore, AB is a tangent. QED. - Tangent from a Point to a circle Property (TPP) IF PA and PB are tangent segments, THEN PA = PB in DAOP, DBOP AO = BO (radii of a common circle) OP is common and OAP = OBP = 90º (TRP) therefore, DAOP DBOP (HS) whence AP = BP, QED.
7 ex. if PA =15, determine the perimeter of DPCD. clearly, DE = DB and EC = CA (TPP) the perimeter of DPCD = PD + DE + PC = PD + DE + EC + PC ( ) + ( CA + PC) = PD + DB = PB + PA = 30 - Tangent Chord Property (TCP) the angle formed by a chord and a tangent is equal to the inscribed angle subtended by the chord in the opposite segment. BAD = BXA ie. BAC = BYA clearly, a = b (ACP), and YBA = 90º (ACP) b + g = 90º (SATT) but g + d = 90º (TRP) whence d = b = a. QED. - Intersecting Chords Property (ICP) IF two chords AB and CD intersect at E THEN AE EB = CE ED AEC = DEB (opposite angles) and ACD = ABD (subtended by common arc) therefore DACE ~ DDBE (AA ~). thus AE ED = CE EB ( ~ Ds) whence AE EB = CE ED (POP) QED.
8 - Intersecting Secants Property (ISP) IF two secants AB and CD meet at P THEN AP PB = CP PD - Corollary to ISP AP PB = ( PT) 2 ex. prove that ADQP is cyclic. clearly, DXBC ~ DXDA and a = g since XQ QB = XP PC = 1 1 therefore QP BC (DST) so b = g (PLT, corresponding angles) thus a = b whence ADQP is cyclic because a and b are subtended by side AP for review questions, see: p. 112 #1-7 p. 108 #1, 2, 4, 5, 7, 9-12, 14-16, 18, 21
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