1. Finding Lengths of Segments in Chords When two chords intersect in the interior of a circle, each chord is divided into two segments which are called segments of a chord. Theorem 10.15 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. THEOREM The following theorem gives a relationship between the lengths of the four segments that are formed. EA EB = EC ED

2. Finding Lengths of Segments in Chords You can use similar triangles to prove Theorem 10.15. PROVE EA EB = EC ED GIVEN AB, CD are chords that intersect at E. Paragraph Proof Draw DB and AC. The lengths of the sides are proportional. EA ED = EC EB Cross Product Property EA EB = EC ED Because  C and  B intercept the same arc,  C   B. Likewise,  A   D. So, the lengths of corresponding sides are proportional. By the AA Similarity Postulate, AEC ~ DEB.

3. Finding Segment Lengths Use Theorem 10.15. RQ RP = RS RT Substitute. RQ RP = RS RT Simplify. 9x = 18 Divide each side by 9. x = 2 63 x 9 Chords ST and PQ intersect inside the circle. Find the value of x.

4. Using Segments of Tangents and Secants In the figure shown below, PS is called a tangent segment because it is tangent to the circle at the endpoint. Similarly, PR is a secant segment and PQ is the external segment of PR.

5. EA EB = EC ED THEOREMS Using Segments of Tangents and Secants Theorem 10.16 If two secant segments share the same endpoint outside a circle, then the product of the length of one secant segment and the length of its external segment equals the product of the length of the other secant segment and the length of its external segment. (EA) 2 = EC ED Theorem 10.17 If a secant segment and a tangent segment share an endpoint outside a circle, then the product of the length of the secant segment and the length of its external segment equals the square of the length of the tangent segment.

6. Finding Segment Lengths Find the value of x. Use Theorem 10.16. RP RQ = RS RT Substitute. RP RQ = RS RT Simplify. 180 = 10x + 100 Divide each side by 10. 8 = x (x + 10)10 (11 + 9) 9 Subtract 100 from each side. 80 = 10x

7. (CB) 2  CE CD Estimating the Radius of a Circle AQUARIUM TANK You are standing at point C, about 8 feet from a circular aquarium tank. The distance from you to a point of tangency on the tank is about 20 feet. Estimate the radius of the tank. Use Theorem 10.17. (CB) 2 = CE CD Substitute. Simplify. 400  16r + 64 21  r (2r + 8) 8 20 2 Subtract 64 from each side. 336  16r SOLUTION You can use Theorem 10.17 to find the radius. So, the radius of the tank is about 21 feet. Divide each side by 16.

8. (BA) 2 = BC BD Finding Segment Lengths Use the figure to find the value of x. Use Theorem 10.17. (BA) 2 = BC BD Substitute. Simplify. 25 = x 2 + 4x Simplify. (x + 4)x 5 25 2 Write in standard form. 0 = x 2 + 4x – 25 SOLUTION Use Quadratic Formula. Use the positive solution, because lengths cannot be negative.  3.39. So, x = –2 + 29 x = –4 ± 4 2 – 4(1)(–25) 2 x = –2 ± 29