Warm-Up a. mBD b. mACE c. mDEB d. mABC 140° 130° 230° 270°

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Presentation transcript:

Warm-Up a. mBD b. mACE c. mDEB d. mABC 140° 130° 230° 270° Find the measure of each arc of circle Z. a. mBD b. mACE c. mDEB d. mABC 140° B A 35° 130° 90° Z C 105° 75° 55° 230° E D 270°

Apply Properties of Chords Section 10 – 3 & Section 10 – 6 Apply Properties of Chords

Theorem 10-3 In the same circle, or in congruent circles, two minor arcs are congruent if and only if their corresponding chords are congruent. B C D A

Theorem 10-4 If one chord is a perpendicular bisector of another chord, then the first chord is a diameter. T S Q P R

Theorem 10-5 If a diameter of a circle is perpendicular to a chord, then the diameter bisects the chord and its arc. T S Q P R

Theorem 10-6 In the same circle, or in congruent circles, two chords are congruent if and only if they are equidistant from the center. C A F G P B D

Theorem 10-14 Segments of Chords Theorem If two chords intersect in the interior of a circle, then the product of the lengths of the segments of one chord is equal to the product of the lengths of the segments of the other chord. C A E B D EA • EB = EC • ED

Example 1 ~ ~ Since AB and CD are = chords, then AB = CD. ~ mCD = 108° In circle R, AB = CD and mAB = 108°. Find mCD. A ~ C Since AB and CD are = chords, then AB = CD. ~ 108° R mCD = 108° B D

Example 2 Use the diagram of Circle C to find the length of BF. Tell what theorem you used. A 10 units 10 B D F C Theorem 10-5: If a diameter is perpendicular to a chord than the diameter bisects the chord and its arc. G

Example 3 Use Theorem 10-6 QR = ST 2x + 6 = 3x – 1 x = 7 In the diagram of Circle P, PV = PW, QR = 2x + 6 and ST = 3x – 1. Find QR. Use Theorem 10-6 R QR = ST V S 2x + 6 = 3x – 1 Q W P x = 7 T QR: 2(7) + 6 = QR = 20 units

Example 4 Use Theorem 10-14 SV • VU = RV • VT x • 4x = (x + 1) • 3x Find RT and SU. Use Theorem 10-14 SV • VU = RV • VT S R x x + 1 T V 3x x • 4x = (x + 1) • 3x 4x2 = 3x2 + 3x 4x x2 = 3x U x = 3 RT = (x + 1) + 3x SU = x + 4x RT = (3 + 1) + 3(3) SU = 3 + 4(3) RT = 13 SU = 15

Homework Section 10-3 Page 667 – 668 (3 – 11, 15, 18 – 20) (3, 4, 13, 16)