Segments of Chords Theorem

Key Concepts

• Find lengths using theorem 10.14.
• Use Theorem 10.15 segments of secants theorem.
• Find lengths using theorem 10.16.
• Solve a real –world problem.

Segments of Chords Theorem

Ok, let’s know about segments of chord. 

The following theorem gives a relationship between the lengths of the four segments that are formed. 

Investigate Segment Lengths 

What is the relationship between the lengths of segments in a circle? 

You can use geometry drawing software to find a relationship between the segments formed by two intersecting chords. 

Draw a circle with two chords. 

Draw Conclusions  

Use your observations to complete these exercises. 

1. What do you notice about the products you found in Step 4?  
2. Drag points A, B, C, and D, keeping point E inside the circle. What do you notice about the new products from Step 4?  
3. Make a conjecture about the relationship between the four-chord segments.  
4. Let PQ¯¯ and RS¯¯ be two chords of a circle that intersect at the point T.  

If PT =9, QT = 5, and RT = 15, use your conjecture from Exercise 3 to find ST. 

PT . TQ = TR . TS 

9 . 5 =15 . TS 

45 = 15 .TS 

TS = 3 

When two chords intersect in the interior of a circle, each chord is divided into two segments that are called segments of the chord. 

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. 

Plan for Proof  

To prove Theorem 10.14, construct two similar triangles. The lengths of the corresponding sides are proportional, so EA /ED = EC /EB. By the Cross Products Property, EA . EB = EC . ED. 

Proof: 

Draw AD and AC. 

Compare △DEA and △ BEC 

∠A ≅ ∠C    Intersect at the same arc 

∠D ≅ ∠D     Intersect at the same arc 

By the AA Similarity postulate  

△DEA ∼△ BEC 

So, the lengths of corresponding sides are proportional. 

ED/EB = EA/EC

EA . EB = EC . ED 

Example 1: 

Find ML and JK 

Solution: 

Find ML and JK. 

NK . NJ = NL . NM                           (Use Theorem 10.14) 

 x . (x + 4) = (x + 1) . (x + 2)             (Substitute)  

x²+ 4x = x² + 3x + 2                        (Simplify) 

4x = 3x + 2                                       (Subtract x_² from each side)  

∴ x = 2                                             (Solve for x) 

Find ML and JK by substitution. 

ML = (x+2)+(x+1) 

        = 2+2+2+1 

ML = 7 

JK = x+(x+4) 

     = 2+2+4 

JK = 8 

Example 2:

Determine which theorem you would use to find x. Then find the value of x. 

Solution: 

By using Theorem 10.14 

PT . TQ =RT . TS 

x.(3) = (4).(6) 

3x = 24     

X = 24/3

x = 8

Example 3: Determine which theorem you would use to find x. Then find  the value of x. 

Solution: 

By Using theorem 10.14 

AB . BC = EB . BD 

x . 18 = 6. 16              (Substitute) 

18 x = 144                   (Simplify) 

x=8 

Tangents And Secants 

A secant segment is a segment that contains a chord of a circle and has exactly one endpoint outside the circle. The part of a secant segment that is outside the circle is called an external segment. 

Theorem 10.15 Segments of Secants Theorem 

If two secant segments share the same endpoint outside a circle, then the product of the lengths of one secant segment and its external segment equals the product of the lengths of the other secant segment and its external segment. 

Proving Theorem 10.15  

Use the plan to prove Theorem 10.15.  

GIVEN: EB and ED are secant segments.  

PROVE: EA . EB = EC . ED  

Plan for Proof: Draw AD and BC. Show that △BCE and △DAE are similar.  

Use the fact that corresponding side lengths in similar triangles are proportional. 

Standardized Test Practice 

1. What is the value of x? 

a. 6 b. 6 2/3 c. 8 d. 9 

Solution: 

RQ . RP = RS . RT       (Use Theorem 10.15) 

4.(5+4) = 3. (x+3)      (Substitute)

4.9 =3x+9                (Simplify) 

36=3x+9 

9 =x                       (Solve for x) 

2. Determine which theorem you would use to find x. Then find the value of x. 

Solution: 

By using segment of secants theorem 10.15 

AB . AC = AE . AF 

3.(3+x+2) = (x+1)(x+1 +x-1) 

3.(5+x)    = (x+1)(2x) 

15+3x     = 2x²+2x 

2x²+2x-3x-15 = 0 

2x²−x−15=0

Measurements are not negative, so leave

x=−2.5

So, x=3 

3. Determine which theorem you would use to find x. Then find the value of x. 

Solution: 

By using segments of secants theorem 10.15 

Theorem 10.16  

Segments of Secants and Tangents Theorem  

If a secant segment and a tangent segment share an endpoint outside a circle, then the product of the lengths of the secant segment and its external segment equals the square of the length of the tangent segment. 

Proving Theorem 10.16  

Use the plan to prove Theorem 10.16.  

GIVEN:

EA is a tangent segment. ED is a secant segment.  

PROVE:

EA² =EC . ED  

Plan for Proof  

Draw

AD and AC

Use the fact that corresponding side lengths in similar triangles are proportional. 

Find lengths using Theorem 10.16 

1. Use the figure at the right to find RS. 

Solution: 

RD2 =RS . RT             (Use Theorem 10.16)

16²  = x.(x+8)            (Substitute) 

256 = x² +8x 

I.e.,  x² + 8x – 256 = 0                          

Compare with a

x²+bx+c=0

a = 1, b = 8 and c =-256 

Use quadratic formula 

x=−b ± √ b²−4ac / 2a

Substitute a, b and c  values  

x = –4+4 √17 (Simplify) 

So, x= –4+4 √17

         = –4 +4 x 4.123 

         = –4 + 16.492 

         = 14.492 

       x ≈ 12.49 

RS≈12.49. 

2. Determine which theorem you would use to find x. Then find the value of x. 

Solution: 

BY  

Theorem 10.16  

Segments of Secants and Tangents Theorem  

PR² = PQ . PR 

12² = x(x+10)                  (Simplify) 

144 = x²+10x 

I.e.,   x²+10x–144=0        (Write standard form)      

x=−10±102−4(1)(−144)

   Use quadratic formula or factor 

Simplify x = 8 

3. Find the value of x. 

Solution: 

PS2 =PQ .PR Use theorem 10.16

Or

PQ .PR = PS2

5.(x+5) = 72

5x+25 = 49 Simplify

5x = 24

x = 24/5

x= 4.8

Science  

1. Tethys, Calypso, and Telesto are three of Saturn’s moons. Each has a nearly circular orbit of 295,000 kilometers in radius. The Cassini-Huygens spacecraft entered Saturn’s orbit in July 2004. Telesto is on the point of tangency. Find the distance DB from Cassini to Tethys. 

Solution: 

Cassini is about 496,494 kilometers from Tethys. 

Let’s check our knowledge  

Answer the following questions

1. Find the value of x in the figure. 

2. Find the value of x in the figure. 

3. Find the value of x in the figure. 

4. Find the value of x in the figure. 

Answers

1. Solution: 

By the Theorem 10.14 

OA x OM = OJ x OL 

12x = 3 x 8  

12x = 24 

     x = 2 

2. Solution: 

By the Theorem 10.14 

OA x OM = OJ x OL 

   6z = 36 

     z = 6 

3. Solution: 

By the Theorem 10.16 

7² = 5(5 + x) 

49 = 25 + 5x  

5x = 49 – 25  

5x = 24 

   x = 24/5

   x = 4.8 

4. Solution: 

By Theorem 10.16  

x(x + 10) = 7² 

x² + 10x = 49 

X² + 10x – 49 = 0 

Exercise

• Chords ST and PQ intersect inside the circle. Find the value of x.

• Find Segments Lengths

• You are standing at point C, about 12 feet from a circular aquarium tank. The distance from you to the point of tangency on the tank is about 30 feet. Estimate the radius of the tank.

• Use the figure at the right to find the value of x.

• Find the value of x.

Concept map 

What we have learned

• Understand how to find lengths using Theorem 10.14.
• Understand how to use Theorem 10.15 secants of secants theorem.
• Understand how to find lengths using theorem 10.16.
• Understand how to solve a real-world problem.