Transcript
Page 1: Lesson 5: Tangent Lines and the Tangent Function

NYS COMMON CORE MATHEMATICS CURRICULUM M4 Lesson 5 PRECALCULUS AND ADVANCED TOPICS

Lesson 5: Tangent Lines and the Tangent Function

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Lesson 5: Tangent Lines and the Tangent Function

Student Outcomes

Students construct a tangent line from a point outside a given circle to the circle (G-C.A.4).

Lesson Notes

This lesson is designed to address standard G-C.A.4, which involves constructing tangent lines to a given circle from a

point outside the circle. The lesson begins by revisiting the geometric origins of the tangent function from Algebra II

Module 2 Lesson 6. Students then explore the standard by means of construction by compass and straightedge and by

paper folding. Students are provided several opportunities to create mathematical arguments to explain their

constructions.

Classwork

Exercises 1โ€“4 (4 minutes)

In this sequence of exercises, students recall the connections between the trigonometric functions and the geometry of

the unit circle.

Exercises 1โ€“12

The circle shown to the right is a unit circle, and the length of ๐‘ซ๏ฟฝฬ‚๏ฟฝ is ๐…

๐Ÿ‘

radians.

1. Which segment in the diagram has length ๐ฌ๐ข๐ง (๐…๐Ÿ‘

)?

The sine of ๐…

๐Ÿ‘ is the vertical component of point ๐‘จ, so ๐‘จ๐‘ฉฬ…ฬ… ฬ…ฬ… has

length ๐ฌ๐ข๐ง (๐…๐Ÿ‘

).

2. Which segment in the diagram has length ๐œ๐จ๐ฌ (๐…๐Ÿ‘

)?

The cosine of ๐…

๐Ÿ‘ is the horizontal component of point ๐‘จ, so ๐‘ถ๐‘ฉฬ…ฬ…ฬ…ฬ…ฬ…

has length ๐œ๐จ๐ฌ (๐…๐Ÿ‘

).

3. Which segment in the diagram has length ๐ญ๐š๐ง (๐…๐Ÿ‘

)?

Since โ–ณ ๐‘ถ๐‘จ๐‘ฉ is similar to โ–ณ ๐‘ถ๐‘ช๐‘ซ, ๐‘ช๐‘ซ

๐Ÿ=

๐‘จ๐‘ฉ

๐‘ถ๐‘ฉ=

๐ฌ๐ข๐ง(๐…๐Ÿ‘

)

๐œ๐จ๐ฌ(๐…๐Ÿ‘

) . Thus, ๐‘ช๐‘ซฬ…ฬ… ฬ…ฬ… has

length ๐ญ๐š๐ง (๐…๐Ÿ‘

).

Page 2: Lesson 5: Tangent Lines and the Tangent Function

NYS COMMON CORE MATHEMATICS CURRICULUM M4 Lesson 5 PRECALCULUS AND ADVANCED TOPICS

Lesson 5: Tangent Lines and the Tangent Function

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4. Which segment in the diagram has length ๐ฌ๐ž๐œ(๐…๐Ÿ‘

)?

Since โ–ณ ๐‘ถ๐‘จ๐‘ฉ is similar to โ–ณ ๐‘ถ๐‘ช๐‘ซ, ๐‘ถ๐‘ซ

๐‘ถ๐‘ฉ=

๐‘ถ๐‘ช

๐‘ถ๐‘จ. Since ๐‘ถ๐‘จ = ๐‘ถ๐‘ซ = ๐Ÿ, we have ๐‘ถ๐‘ช =

๐Ÿ๐‘ถ๐‘ฉ

=๐Ÿ

๐œ๐จ๐ฌ(๐…๐Ÿ‘

). Thus, ๐‘ถ๐‘ชฬ…ฬ… ฬ…ฬ… has

length ๐ฌ๐ž๐œ(๐…๐Ÿ‘

).

Discussion (7 minutes): Connections Between Geometry and Trigonometry

Do you recall why the length of ๐ถ๐ทฬ…ฬ… ฬ…ฬ… is called the tangent of ๐œ‹

3? And do you recall why the length of ๐‘‚๐ถฬ…ฬ… ฬ…ฬ… is called

the secant of ๐œ‹

3? (Hint: Look at how lines containing ๐ถ๐ทฬ…ฬ… ฬ…ฬ… and ๐‘‚๐ถฬ…ฬ… ฬ…ฬ… are related to the circle.)

The line containing ๐ถ๐ทฬ…ฬ… ฬ…ฬ… is tangent to the circle because it intersects the circle once, so it makes sense to

refer to ๐ถ๐ทฬ…ฬ… ฬ…ฬ… as a tangent segment and to refer to the length ๐ถ๐ท as the tangent of ๐œ‹

3.

The line containing ๐‘‚๐ถฬ…ฬ… ฬ…ฬ… is a secant line because it intersects the circle twice, so it makes sense to refer to

๐‘‚๐ถฬ…ฬ… ฬ…ฬ… as a secant segment and to refer to the length ๐‘‚๐ถ as the secant of ๐œ‹

3.

How can you be sure that the line containing ๐ถ๐ทฬ…ฬ… ฬ…ฬ… is, in fact, a tangent line?

We see that ๐ถ๐ท โƒก is perpendicular to segment ๐‘‚๐ทฬ…ฬ… ฬ…ฬ… at point ๐ท, and so ๐ถ๐ท โƒก must be tangent to the circle.

Explain how this diagram can be used to show that tan (๐œ‹

3) =

sin(๐œ‹3)

cos(๐œ‹3)

.

The triangles in the diagram have two pairs of congruent angles, so they are similar to each other. It

follows that their sides are proportional, and so tan(

๐œ‹

3)

sin(๐œ‹

3)

=1

cos(๐œ‹

3). If we multiply both sides by sin (

๐œ‹3),

we obtain the desired result.

MP.7

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Lesson 5: Tangent Lines and the Tangent Function

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Explain how this diagram can be used to show that sin2 (๐œ‹3) + cos2 (

๐œ‹3) = 1.

If we apply the Pythagorean theorem to โ–ณ ๐ด๐‘‚๐ต, we find that (๐ด๐ต)2 + (๐‘‚๐ต)2 = (๐‘‚๐ด)2, so sin2 (๐œ‹3) +

cos2 (๐œ‹3) = 1.

Which trigonometric identity can be established by applying the Pythagorean theorem to โ–ณ ๐ถ๐‘‚๐ท?

tan2 (๐œ‹3) + 12 = sec2 (

๐œ‹3)

What is the scale factor that relates the sides of โ–ณ ๐ด๐‘‚๐ต to those of โ–ณ ๐ถ๐‘‚๐ท?

Point ๐ต is the midpoint of ๐‘‚๐ทฬ…ฬ… ฬ…ฬ… , so the sides of โ–ณ ๐ถ๐‘‚๐ท are twice the length of the corresponding sides of

โ–ณ ๐ด๐‘‚๐ต.

Explain how the diagram above can be used to show that tan (๐œ‹3) = โˆš3.

Since the length of ๐ท๏ฟฝฬ‚๏ฟฝ is ๐œ‹

3 radians, it follows that โ–ณ ๐ถ๐‘‚๐ธ is equilateral, and so its sides must each be 2

units long. Applying the Pythagorean theorem to โ–ณ ๐ถ๐‘‚๐ท, we find 12 + ๐ถ๐ท2 = 22, which means that

๐ถ๐ท2 = 3, and so ๐ถ๐ท = โˆš3. Thus, tan (๐œ‹3) = โˆš3.

This diagram contains a host of information about trigonometry. But can we take this diagram even further?

Letโ€™s press on to some new territory. Much of the remaining discussion is devoted to the topic of

constructions. Get ready to have some fun with paper folding and your compass and straightedge!

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Exploratory Challenge 1 (7 minutes): Constructing Tangents via Paper-Folding

Earlier we observed that ๐ถ๐ท โƒก is tangent to the circle. Can you visualize another line through point ๐ถ that is also

tangent to the circle? Try to draw a second tangent line through point ๐ถ, and label its point of intersection

with the circle as point ๐น.

First, letโ€™s examine this diagram through the lens of transformations. Can you see how to map ๐ถ๐ท โƒก onto ๐ถ๐น โƒก ?

It appears that you could map ๐ถ๐ท โƒก onto ๐ถ๐น โƒก by reflecting it across ๐ถ๐‘‚ โƒก .

Letโ€™s get some hands-on experience with this by performing the reflection in the most natural way

imaginableโ€”by folding a piece of paper.

Give students an unlined piece of copy paper or patty paper, a ruler, and a compass.

The goal of this next activity is essentially to recreate the diagram above. Students take one tangent line and use a

reflection to produce a second tangent line.

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NYS COMMON CORE MATHEMATICS CURRICULUM M4 Lesson 5 PRECALCULUS AND ADVANCED TOPICS

Lesson 5: Tangent Lines and the Tangent Function

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Choose a point in the middle of your paper, and label it ๐‘‚. Use your compass to draw a circle with center ๐‘‚.

Choose a point on that circle, and label it ๐ท. Use your straightedge to draw the line through ๐‘‚ and ๐ท,

extending it beyond the circle. Can you see how to fold the paper in such a way that the crease is tangent to

the circle at point ๐ท? Think about this for a moment.

We want to create a line that is perpendicular to ๐‘‚๐ท โƒก at point ๐ท, so we fold the paper in such a way that

the crease is on point ๐ท and ๐‘‚๐ท โƒก maps to itself.

Now choose a point on the crease, and label it ๐ถ. Try to create a second tangent line through point ๐ถ. Can you

see how to do this?

We just need to fold the paper so that the crease is on points ๐ถ and ๐‘‚. We can see through the back of

the paper where point ๐ท is, and this is where we mark a new point ๐น, which is the other point of

tangency. In other words, ๐ถ๐น โƒก is also tangent to the circle.

Do you think you could construct an argument that ๐ถ๐น โƒก is truly tangent to the circle? Think about this for a few

minutes, and then share your thoughts with the students around you.

A reflection is a rigid motion, preserving both distances and angles. Since ๐ถ๐ท โƒก is tangent to the circle,

we know that โˆ ๐ถ๐ท๐‘‚ is a right angle. And since the distance from ๐‘‚ to ๐ท is preserved by the reflection,

it follows that point ๐น, which is the image of ๐ท, is also located on the circle because ๐‘‚๐น = ๐‘‚๐ท, both of

which are radii of the circle. This means that ๐น is on the circle and that โˆ ๐ถ๐น๐‘‚ is a right angle, which

proves that ๐ถ๐น โƒก is indeed tangent to the circle.

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NYS COMMON CORE MATHEMATICS CURRICULUM M4 Lesson 5 PRECALCULUS AND ADVANCED TOPICS

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Exploratory Challenge 2 (5 minutes): Constructing Tangents Using a Compass

Letโ€™s try to produce this tangent line without using paper-folding. Can you see how to produce this diagram

using only a compass and a straightedge? Take a few minutes to explore this problem.

We use a straightedge to draw a line that is perpendicular to ๐‘‚๐ท โƒก at point ๐ท and choose a point ๐ถ on

this perpendicular line as before. Next, we use a compass to draw a circle around point ๐ถ that contains

point ๐ท. This circle will intersect the original circle at a point ๐น, which is the desired point of tangency.

Now make an argument that your construction produces a line that is truly tangent to the circle.

Each point on the circle around point ๐‘‚ is the same distance from ๐‘‚, so ๐‘‚๐น = ๐‘‚๐ท. Also, each point on

the circle around point ๐ถ is the same distance from point ๐ถ, so ๐ถ๐น = ๐ถ๐ท. Since ๐ถ๐‘‚ = ๐ถ๐‘‚,

โ–ณ ๐‘‚๐ถ๐ท โ‰…โ–ณ ๐‘‚๐ถ๐น by the SSS criterion for triangle congruence. Since โˆ ๐ถ๐ท๐‘‚ is a right angle, it follows by

CPCTC that โˆ ๐ถ๐น๐‘‚ is also a right angle. This proves that ๐ถ๐น โƒก is indeed tangent to the circle.

Exploratory Challenge 3 (7 minutes): Constructing the Tangents from an External Point

Now letโ€™s change the problem slightly. Suppose we are given a circle and an external point. How might we go

about constructing the lines that are tangent to the given circle and pass through the given point? Take

several minutes to wrestle with this problem, and then share your thoughts with a neighbor.

MP.1

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This is a tough problem, isnโ€™t it? Letโ€™s see if we can solve it by using some logical reasoning. What do you

know about the lines you want to create?

We want the lines to be tangent to the circle, so they must be constructed in such a way that โˆ ๐ถ๐น๐‘‚ and

โˆ ๐ถ๐ท๐‘‚ are right angles.

Perhaps we can use this to our advantage. Letโ€™s consider the entire locus of points ๐น such that โˆ ๐ถ๐น๐‘‚ is a right

angle. Can you describe this locus? Can you construct it? This is a challenge in its own right. Letโ€™s explore this

challenge using a piece of paper, which comes with some built-in right angles.

Take a blank piece of paper, and mark two points several inches apart; label them ๐ด and ๐ต. Now take a second

piece of paper, and line up one edge on ๐ด and the adjacent edge on ๐ต. Mark the point where the corner of the

paper is, then shift the corner to a new spot, keeping the edges against ๐ด and ๐ต. Quickly repeat this 20 or so

times until a clear pattern emerges. Then describe what you see.

This locus of points is a circle with diameter ๐ด๐ตฬ…ฬ… ฬ…ฬ… .

Now construct this circle using your compass.

We just need to construct the midpoint of ๐ด๐ตฬ…ฬ… ฬ…ฬ… , which is the center of the circle.

We are ready to apply this result to our problem involving tangents to a circle. Do you see how this relates to

the problem of creating a tangent line?

By drawing a circle with diameter ๐‘‚๐ถฬ…ฬ… ฬ…ฬ… , we create two points ๐น and ๐ท with two important properties.

First, they are both on the circle around ๐‘‚. Second, they are on the circle with diameter ๐‘‚๐ถฬ…ฬ… ฬ…ฬ… , which

means that โˆ ๐‘‚๐น๐ถ and โˆ ๐‘‚๐ท๐ถ are right angles, just as we require for tangent lines.

D

F

O

C

Page 8: Lesson 5: Tangent Lines and the Tangent Function

NYS COMMON CORE MATHEMATICS CURRICULUM M4 Lesson 5 PRECALCULUS AND ADVANCED TOPICS

Lesson 5: Tangent Lines and the Tangent Function

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Exercises 5โ€“7 (4 minutes)

Instruct students to perform the following exercises and to compare their results with a partner.

5. Use a compass to construct the tangent lines to the given circle that pass through the given point.

Sample solution:

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NYS COMMON CORE MATHEMATICS CURRICULUM M4 Lesson 5 PRECALCULUS AND ADVANCED TOPICS

Lesson 5: Tangent Lines and the Tangent Function

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6. Analyze the construction shown below. Argue that the lines shown are tangent to the circle with center ๐‘ฉ.

Since point ๐‘ซ is on circle ๐‘ช, we know that โˆ ๐‘ฉ๐‘ซ๐‘จ is a right angle. Since point ๐‘ซ is on circle ๐‘ฉ and โˆ ๐‘ฉ๐‘ซ๐‘จ is a right

angle, it follows that ๐‘จ๐‘ซ โƒก is tangent to circle ๐‘ฉ. In the same way, we can show that ๐‘จ๐‘ฌ โƒก is tangent to circle ๐‘ฉ.

7. Use a compass to construct a line that is tangent to the circle below at point ๐‘ญ. Then choose a point ๐‘ฎ on the

tangent line, and construct another tangent to the circle through ๐‘ฎ.

Sample solution:

F

O

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Exercises 8โ€“12 (5 minutes)

Instruct students to perform the following exercises and to compare their results with a partner.

8. The circles shown below are unit circles, and the length of ๐‘ซ๏ฟฝฬ‚๏ฟฝ is ๐…

๐Ÿ‘ radians.

Which trigonometric function corresponds to the length of ๐‘ฌ๐‘ญฬ…ฬ… ฬ…ฬ… ?

The length of ๐‘ฌ๐‘ญฬ…ฬ… ฬ…ฬ… represents the cotangent of ๐…

๐Ÿ‘.

9. Which trigonometric function corresponds to the length of ๐‘ถ๐‘ญฬ…ฬ… ฬ…ฬ… ?

The length of ๐‘ถ๐‘ญฬ…ฬ… ฬ…ฬ… represents the cosecant of ๐…

๐Ÿ‘.

10. Which trigonometric identity gives the relationship between the lengths of the sides of โ–ณ ๐‘ถ๐‘ฌ๐‘ญ?

๐œ๐จ๐ญ๐Ÿ (๐…

๐Ÿ‘) + ๐Ÿ๐Ÿ = ๐œ๐ฌ๐œ๐Ÿ (

๐…

๐Ÿ‘)

11. Which trigonometric identities give the relationships between the corresponding sides of โ–ณ ๐‘ถ๐‘ฌ๐‘ญ and โ–ณ ๐‘ถ๐‘ฎ๐‘จ?

We have ๐Ÿ

๐ฌ๐ข๐ง(๐…

๐Ÿ‘)

=๐œ๐จ๐ญ(

๐…

๐Ÿ‘)

๐œ๐จ๐ฌ(๐…

๐Ÿ‘)

and ๐Ÿ

๐ฌ๐ข๐ง(๐…

๐Ÿ‘)

=๐œ๐ฌ๐œ(

๐…

๐Ÿ‘)

๐Ÿ.

12. What is the value of ๐œ๐ฌ๐œ (๐…๐Ÿ‘

)? What is the value of ๐œ๐จ๐ญ (๐…๐Ÿ‘

)? Use the Pythagorean theorem to support your

answers.

Let ๐’™ = ๐‘ฌ๐‘ญ. Then we have ๐’™๐Ÿ + ๐Ÿ๐Ÿ = (๐Ÿ๐’™)๐Ÿ = ๐Ÿ’๐’™๐Ÿ, which means ๐Ÿ‘๐’™๐Ÿ = ๐Ÿ; therefore, ๐’™ = โˆš๐Ÿ๐Ÿ‘

. So ๐œ๐จ๐ญ (๐…๐Ÿ‘

) = โˆš๐Ÿ๐Ÿ‘

and ๐œ๐ฌ๐œ (๐…๐Ÿ‘

) = ๐Ÿโˆš๐Ÿ๐Ÿ‘

.

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Closing (2 minutes)

If you are given a circle and an external point, how do you use a compass to construct the lines that pass

through the given point and are tangent to the given circle? Use your notebook to write a summary of the

main steps involved in this construction.

Letโ€™s say the center of the circle is point ๐ด and the external point is ๐ต. First, we need to construct the

midpoint of the segment joining ๐ด and ๐ตโ€”call this point ๐ถ. Next, we draw a circle around point ๐ถ that

goes through ๐ด and ๐ต. The points where this circle intersects the first circle are the points of tangency,

so we just connect these points to point ๐ต to form the tangents.

Exit Ticket (4 minutes)

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Name Date

Lesson 5: Tangent Lines and the Tangent Function

Exit Ticket

1. Use a compass and a straightedge to construct the tangent lines to the given circle that pass through the given

point.

2. Explain why your construction produces lines that are indeed tangent to the given circle.

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Exit Ticket Sample Solutions

1. Use a compass and a straightedge to construct the tangent lines to the given circle that pass through the given point.

2. Explain why your construction produces lines that are indeed tangent to the given circle.

Since points ๐‘ซ and ๐‘ฌ are on circle ๐‘ช, โˆ ๐‘ฉ๐‘ซ๐‘จ and โˆ ๐‘ฉ๐‘ฌ๐‘จ are right angles. Thus, ๐‘จ๐‘ซ โƒก and ๐‘จ๐‘ฌ โƒก are tangent to circle ๐‘ฉ.

Problem Set Sample Solutions

1. Prove Thalesโ€™ theorem: If ๐‘จ, ๐‘ฉ, and ๐‘ท are points on a circle where ๐‘จ๐‘ฉฬ…ฬ… ฬ…ฬ… is a diameter of the circle, then โˆ ๐‘จ๐‘ท๐‘ฉ is a

right angle.

Since ๐‘ถ๐‘จ = ๐‘ถ๐‘ท = ๐‘ถ๐‘ฉ, โ–ณ ๐‘ถ๐‘ท๐‘จ and โ–ณ ๐‘ถ๐‘ท๐‘ฉ are isosceles triangles. Therefore,

๐’Žโˆ ๐‘ถ๐‘จ๐‘ท = ๐’Žโˆ ๐‘ถ๐‘ท๐‘จ, and ๐’Žโˆ ๐‘ถ๐‘ท๐‘ฉ = ๐’Žโˆ ๐‘ถ๐‘ฉ๐‘ท.

Let ๐’Žโˆ ๐‘ถ๐‘ท๐‘จ = ๐œถ and ๐’Žโˆ ๐‘ถ๐‘ท๐‘ฉ = ๐œท. The sum of three internal angles of โ–ณ ๐‘จ๐‘ท๐‘ฉ

equals ๐Ÿ๐Ÿ–๐ŸŽยฐ.

Therefore, ๐œถ + (๐œถ + ๐œท) + ๐œท = ๐Ÿ๐Ÿ–๐ŸŽยฐ, so ๐Ÿ๐œถ + ๐Ÿ๐œท = ๐Ÿ๐Ÿ–๐ŸŽยฐ, and ๐œถ + ๐œท = ๐Ÿ—๐ŸŽยฐ. Since

๐’Žโˆ ๐‘จ๐‘ท๐‘ฉ = ๐œถ + ๐œท, we have ๐’Žโˆ ๐‘จ๐‘ท๐‘ฉ = ๐Ÿ—๐ŸŽยฐ, so โˆ ๐‘จ๐‘ท๐‘ฉ is a right angle.

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2. Prove the converse of Thalesโ€™ theorem: If ๐‘จ๐‘ฉฬ…ฬ… ฬ…ฬ… is a diameter of a circle and ๐‘ท is a point so that โˆ ๐‘จ๐‘ท๐‘ฉ is a right angle,

then ๐‘ท lies on the circle for which ๐‘จ๐‘ฉฬ…ฬ… ฬ…ฬ… is a diameter.

Construct the right triangle, โ–ณ ๐‘จ๐‘ท๐‘ฉ.

Construct the line ๐’‰ that is parallel to ๐‘ท๐‘ฉฬ…ฬ… ฬ…ฬ… through point ๐‘จ.

Construct the line ๐’ˆ that is parallel to ๐‘จ๐‘ทฬ…ฬ… ฬ…ฬ… through point ๐‘ฉ.

Let ๐‘ช be the intersection of lines ๐’‰ and ๐’ˆ.

The quadrilateral ๐‘จ๐‘ช๐‘ฉ๐‘ท forms a parallelogram by construction.

By the properties of parallelograms, the adjacent angles are supplementary. Since โˆ ๐‘จ๐‘ท๐‘ฉ is a right angle, it follows

that angles โˆ ๐‘ช๐‘จ๐‘ท, โˆ ๐‘ฉ๐‘ช๐‘จ, and โˆ ๐‘ท๐‘ฉ๐‘ช are also right angles. Therefore, the quadrilateral ๐‘จ๐‘ช๐‘ฉ๐‘ท is a rectangle.

Let ๐‘ถ be the intersection of the diagonals ๐‘จ๐‘ฉฬ…ฬ… ฬ…ฬ… and ๐‘ช๐‘ทฬ…ฬ… ฬ…ฬ… . Then, by the properties of parallelograms, point ๐‘ถ is the

midpoint of ๐‘จ๐‘ฉฬ…ฬ… ฬ…ฬ… and ๐‘ช๐‘ทฬ…ฬ… ฬ…ฬ… , so ๐‘ถ๐‘จ = ๐‘ถ๐‘ฉ = ๐‘ถ๐‘ช = ๐‘ถ๐‘ท. Therefore, ๐‘ถ is the center of the circumscribing circle, and the

hypotenuse of โ–ณ ๐‘จ๐‘ท๐‘ฉ, ๐‘จ๐‘ฉฬ…ฬ… ฬ…ฬ… , is a diameter of the circle.

3. Construct the tangent lines from point ๐‘ท to the circle given below.

Mark any three points ๐‘จ, ๐‘ฉ, and ๐‘ช on the circle, and construct perpendicular bisectors of ๐‘จ๐‘ฉฬ…ฬ… ฬ…ฬ… and ๐‘ฉ๐‘ชฬ…ฬ… ฬ…ฬ… .

Let ๐‘ถ be the intersection of the two perpendicular bisectors.

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Construct the midpoint ๐‘ฏ of ๐‘ถ๐‘ทฬ…ฬ… ฬ…ฬ… .

Construct a circle with center ๐‘ฏ and radius ๐‘ถ๐‘ฏ.

The circle centered at ๐‘ฏ will intersect the original circle ๐‘ถ at points ๐‘จ and ๐‘ฉ.

Construct two tangent lines ๐‘ท๐‘จ โƒก and ๐‘ท๐‘ฉ โƒก .

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4. Prove that if segments from a point ๐‘ท are tangent to a circle at points ๐‘จ and ๐‘ฉ, then ๐‘ท๐‘จฬ…ฬ… ฬ…ฬ… โ‰… ๐‘ท๐‘ฉฬ…ฬ… ฬ…ฬ… .

Let ๐‘ท be a point outside of a circle with center ๐‘ถ, and let ๐‘จ and ๐‘ฉ be points on the circle so that ๐‘ท๐‘จฬ…ฬ… ฬ…ฬ… and ๐‘ท๐‘ฉฬ…ฬ… ฬ…ฬ… are

tangent to the circle. Then, ๐‘ถ๐‘จ = ๐‘ถ๐‘ฉ, ๐‘ถ๐‘ท = ๐‘ถ๐‘ท, and ๐’Žโˆ ๐‘ถ๐‘จ๐‘ท = ๐’Žโˆ ๐‘ถ๐‘ฉ๐‘ท = ๐Ÿ—๐ŸŽยฐ, so โ–ณ ๐‘ท๐‘จ๐‘ถ โ‰… โ–ณ ๐‘ท๐‘ฉ๐‘ถ by the

Hypotenuse Leg congruence criterion. Therefore, ๐‘ท๐‘จฬ…ฬ… ฬ…ฬ… โ‰… ๐‘ท๐‘ฉฬ…ฬ… ฬ…ฬ… because corresponding parts of congruent triangles are

congruent.

5. Given points ๐‘จ, ๐‘ฉ, and ๐‘ช so that ๐‘จ๐‘ฉ = ๐‘จ๐‘ช, construct a circle so that ๐‘จ๐‘ฉฬ…ฬ… ฬ…ฬ… is tangent to the circle at ๐‘ฉ and ๐‘จ๐‘ชฬ…ฬ… ฬ…ฬ… is

tangent to the circle at ๐‘ช.

Construct a perpendicular bisector of ๐‘จ๐‘ฉฬ…ฬ… ฬ…ฬ… .

Construct a perpendicular bisector of ๐‘จ๐‘ชฬ…ฬ… ฬ…ฬ… .

The perpendicular bisectors will intersect at point ๐‘ฏ.

Construct a line through points ๐‘จ and ๐‘ฏ.

Construct a circle with center ๐‘ฏ and radius ๐‘ฏ๐‘จฬ…ฬ…ฬ…ฬ…ฬ….

The circle centered at ๐‘ฏ will intersect ๐‘ฏ๐‘จ โƒก at ๐‘ฐ.

Construct a circle centered at ๐‘ฐ with radius ๐‘ฐ๐‘ฉฬ…ฬ…ฬ…ฬ… .


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