Topics
13. Circles
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United States of America
Grades 9-12

13.06 Lengths in intersecting chords, secants, and tangents

Lesson
Practice

Introduction

In the previous lessons, most of the theorems we learned about chords, secants, and tangents were related to the angles they formed. This lesson will focus on the relationships found betweeen the lengths of the line segments.

Lengths in intersecting chords, secants, and tangents

The following two theorems relate to the lengths of segments formed by secants and tangents to a circle from a common external point:

Secant secant theorem

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 is equal to the product of the length of the other secant segment and the length of its external segment

A circle with two segments that are secants to the circle, and intersecting at a point S outside the circle. One of secants is divided into segments of length m, and length n. The segment of length m is a chord of the circle, and the segment with length n is an external segment. The other secant is divided into segments of length a, and length b. The segment of length a is a chord of the circle, and the segment of length b is an external segment.

For the diagram shown, the secant secant theorem says that\left(n + m\right)n = \left(b+a\right)b

Tangent secant theorem

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 is equal to the square of the length of the tangent segment

A circle with a segment that is secant to the circle, and another segment that is tangent to the circle. The secant and the tangent intersect at a point T outside the circle. The tangent segment has a length of x, and has endpoints at T, and at the point of tangency. The secant has an endpoints at T, and at a point on the circle. It intersects the circle at two points. The secant is divided into two segments of length y, and length z. The segment with length y is a chord, and the segment with length z is an external segment.

For the diagram shown, the tangent secant theorem says thatx^2=\left(z+y\right)z

Intersecting chord 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

A circle with points A, G, B, and F placed clockwise on the circle. Chords A B and F G are drawn, and intersect at a point P in the circle.

For the diagram shown, the intersecting chord theorem says that:

{AP} \cdot {PB} = {FP}\cdot {PG}

Exploration

At the bottom of the applet, click the double right arrow to move through the slides.

  1. Use the applet to prove the intersecting chord theorem by determining a justification for each slide.
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Examples

Example 1

Prove the intersecting chord 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.

Worked Solution
Create a strategy

Using a particular diagram, we are trying to prove that AE \cdot BE = CE \cdot DE.

By constructing \overline{BD} and \overline{AC}, we can create triangles and prove the theorem with properties of similarity.

Apply the idea

We will annotate a diagram as we go through the reasonings.

Given \overline{AB} and \overline{CD} are chords.

\angle ABD \cong \angle ACD by the congruent inscribed angle theorem as they are both angles that subtend the same arc.

\angle BDC \cong \angle BAC by the congruent inscribed angle theorem as they are both angles that subtend the same arc.

\angle BED \cong \angle AEC by the vertical angles theorem.

\triangle BED \sim \triangle CEA by Angle-Angle similarity.

Since \triangle BED \sim \triangle CEA we have that: \dfrac{b}{d}=\dfrac{c}{a} rearranging this we get: b \cdot a = c \cdot d as required.

Example 2

Given AB = 5, CB = 3 and CD = 2x+5. Solve for x.

A circle and with points D, C, and A placed clockwise on the circle. Secant ray B D passing through C is drawn. C is between B and D. A ray from B to A is drawn. B A is tangent to the circle at A.
Worked Solution
Create a strategy

We have been given lengths of all the segments needed for the tangent secant theorem, so we can use that to construct an equation and solve for x.

Apply the idea
\displaystyle AB^2\displaystyle =\displaystyle \left(BC + CD \right) BCFormula for tangent secant theorem
\displaystyle 5^2\displaystyle =\displaystyle \left[3+(2x+5)\right]\cdot \left(3\right)Substitution
\displaystyle 25\displaystyle =\displaystyle \left(2x+8\right)3Combine like terms
\displaystyle 25\displaystyle =\displaystyle 6x+24Distribute the 3
\displaystyle 1\displaystyle =\displaystyle 6xSubtract 24 from both sides
\displaystyle \dfrac{1}{6}\displaystyle =\displaystyle xDivide both sides by 6
\displaystyle x\displaystyle =\displaystyle \dfrac{1}{6}Symmetric property of equality

Therefore, x=\dfrac{1}{6}.

Example 3

Given DE = 20, CD = 11 and BC = 13. Find the length of the segment AB.

A circle with points D, E, A, and B are on the circle. Secant segment C E which passes through D, and secant segment C A which passes through B are drawn.
Worked Solution
Create a strategy

We have been given lengths of all the segments needed for the secant secant theorem, so we can use that to construct an equation and solve for x.

Apply the idea
\displaystyle \left(DE + CD\right)CD\displaystyle =\displaystyle \left(BC + AB\right)BCFormula for secant secant theorem
\displaystyle \left(20 + 11\right)11 \displaystyle =\displaystyle \left(13+AB\right)13Substitution
\displaystyle 341\displaystyle =\displaystyle \left(13+AB\right)13Combine like terms and simplify
\displaystyle 341\displaystyle =\displaystyle 169 + 13 \cdot ABDistribute the 13
\displaystyle 172\displaystyle =\displaystyle 13 \cdot ABSubtract 169 from both sides
\displaystyle \dfrac{172}{13}\displaystyle =\displaystyle ABDivide both sides by 13
\displaystyle AB\displaystyle =\displaystyle \dfrac{172}{13}Symmetric property of equality

Therefore, AB=\dfrac{172}{13}.

Example 4

Determine the length of \overline{AC} if AB = 7 \text{ cm}, AD = AE, and AE = 3 \text{ cm}.

A circle with points C, E, D, and B placed clockwise on the circle. Chords C D and B E are drawn, and intersect at a point A in the circle.
Worked Solution
Create a strategy

We can use the intersecting chord theorem. We also know that AD = 3 \text{ cm} since AD = AE.

Apply the idea
\displaystyle AB \cdot AE \displaystyle =\displaystyle AD \cdot ACIntersecting chord theorem
\displaystyle 7 \cdot 3\displaystyle =\displaystyle 3 \cdot ACSubstitute known values
\displaystyle 7\displaystyle =\displaystyle ACDivide both sides by 3

The length of AC is 7 \text{ cm}.

Reflect and check

Since we know that AE = AD, then the intersecting chord theorem tells us that AC = AB as well. In this case, both AC and AB measure 7 \text{ cm}.

Idea summary

When two secants intersect at a point exterior to a circle, the lengths of the secants and the external parts are proportional. When two chords intersect at a point interior to a circle, the chords are divided proportionally.

When a secant and a tangent segment intersect at a point exterior to the circle, the product of the length of the secant segment and the length of its external segment is equal to the square of the length of the tangent segment.

Outcomes

G.C.A.2

Identify and describe relationships among inscribed angles, radii, and chords. Include the relationship between central, inscribed, and circumscribed angles; inscribed angles on a diameter are right angles; the radius of a circle is perpendicular to the tangent where the radius intersects the circle.

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