1) With a protractor (or using CABRI), carefully measure nacb and write down your result.


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1 4.5 The Circle Theorem Moment for Discovery: Inscribed Angles Draw a large circle and any of its chords AB, as shown. Locate three points C, C', and C'' at random on the circle and on the same side of the chord. 1) With a protractor (or using CABRI), carefully measure nacb and write down your result. 2) Repeat Step1 for nac'b and nac''b. 3) What did you discover? 4) If D and D' are points on the circle on the opposite side of chord AB from C, what do you think must be true of mnadb and mnad'b? Verify by direct measurement (or by using CABRI). 5. Did you make any discoveries? Solution 3. The discovery should be that mnc = mnc' = mnc''. 4. mnd = mnd' 5. mnd = mnc. Theorem 1: The measure of an inscribed angle of a circle equals onehalf that of its intercepted arc.
2 Proof: We must show that in the figure, m ABC = mac in all three cases. n Case I: When center O lies on side BC of naob. 1) Since BOC, naoc is an exterior angle of AOB. Hence, ) = x + y (Euclidean Exterior Angle Theorem, 4.1) 2) OA and OB are radii of circle O, so OA = OB and x = y. (Isosceles Triangle Theorem) 3) ) = 2x or x = ) (algebra) 4) ¾ mn ABC = x = ) = mac (definition of circular measure) Case II: When center O lies interior to nabc 1) Consider the diameter BD passing through B. Then mnabc = mnabd + mndbc (Angle Addition Axiom) 2) ¾ mn ABC = mad + mdc = mac (by Case 1 and Additivity of Circular Arc Measure) Case III: When center O lies exterior to nabc 1) Let BD be the diameter through B and construct the tangent to circle O at B; the circle and points A, D, and C all lie on the same side of the tangent line BE Ç, and, using coordinates for the rays p p p p p through B and Theorem 1 of 2.5, either BA  BD  BC, BA  BC  p p p p BD, or BC  BA  BD. Since O is not in the interior of nabc, we
3 p p p rule out the case BA  BC  BD. Then mnabc = mnabd  mncbd p p p 2) Assume, therefore, that BA  BC  BD. Then mnabc = mnabd  mncbd. 3) m ABC = Arc Measure) n mad  mcd = mac (Case 1 and Additivity of Corollary: An angle whose vertex lies inside a circle and whose sides are intersecting chords of the circle (intercepting arcs of measure x and y) has measure ) = (x + y). Proof: mn BAC = ) + 9 = x + y (Euclidean Exterior Angle Theorem) Corollary: An angle whose vertex is exterior to a circle and whose sides are intersecting secants fo the circle (intercepting arcs of measure x and y) has measure ) = l x  y l. Proof: ) = mnbac + 9 Ê x = mnbac + y, or mnbac = (x  y)
4 Corollary: An angle formed by a chord and tangent of a circle, with its vertex at the point of tangency and intercepting an arc of measure x on that cirlce, has measure ) = x. Ç Ç Proof: Since BD ² AC, ) = m n BAC (Property Z). Since the Ç radius through A is perpendicular to line AC (Tangent Theorem), and AE Ç Ç ¼ BD, then AE Ç ¼ AC, Ç and AC Ç passes through the center of the circle. Hence AE bisects chord BD and E is the midpoint of BD. Then AD = AB and n D = ) (Isosceles Triangle Theorem). ¾ mnbac = ) = mnd = x. Moment for Discovery: The Two Cord Theorem Point P is an interior point of a circle having center O and radius È $%, as shown. In addition, OP is the base of an isosceles right triangle. Chord AB is determined, which contains a leg of this isosceles right trangle. 1. Determine, by the Pythagorean Theorem, the values for AP and PB. (Did you get AP = 2?)
5 2. A second chord CD through P is perpendicular to OP (second diagram). Find the values of CP and DP. (Hint: P is the midpoint of CD in this case.) 3. A third chord EF passes through P and makes an angle of measure 60 with OP. Find the values EP and FP, using trigonometry. (Hint: Let Q' be the midpoint of EF, and let u, v be the sides of the right triangle in the third diagram; solve u, v, $ and EQ. You should get u = È '.) 4. Find EP = EQ + v and PF = EQ  v. 5. Check all calculations before you proceed. 6. Evaluate the three products AP PB, CP PD, and EP PF. Did anything unusual happen? Solution 1. QB = ÉÐ È$%Ñ * = È& = 5 = QA. ¾ AP = 2 and PB = PC = PD = È È $% ) œ ' œ %Þ 3. OP = 3È , v ¾ u È)Þ!!! %Þ&!! ; EQ' È $ ( $% $Þ& (Exact values: v œ, u = É, and EQ' = ( È )Ñ È 4. EP and PF (Exact values are ( È) $ È )
6 6. AP PB = 2 8 = 16, CP PD = 4 4 = 16, EP PF (Discovery: AP PB = CP PD = EP PF) Theorem 2: TwoChord Theorem When two chords of a circle intersect, the product of the lengths of the segments formed on one chord equals that on the other chord. That is, AP PB = CP PD Proof: Let P be the point of intersection of chords AB and CD of the circle. 1) mn1 = mn2 (Inscribed Angle Theoremboth angles subtend the same arc AQD) 2) mn3 = mn4 (Vertical Pair Theorem) 3) APC µ DPB (AA Similarity Criterion) 4) PA/PD = PC/PB ¾ AP PB = CP PD (ratios of corresponding sides of similar triangles are equal) Theorem 3: Secant Theorem Ç Ç If a secant PA and tangent PC meet a circle at the respective points A, B, and C (point of contact), then PC = PA PB
7 Corollary: Two Secant Theorem Ç Ç If two secant PA and PC of a circle meet the circle at A, B, C, and D, respectively, then PA PB = PC PD. Proof: Draw a tangent from P and apply the Secant/Tangent Theorem to both secants.
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