NYS COMMON CORE MATHEMATICS CURRICULUM M4 Lesson 5 PRECALCULUS AND ADVANCED TOPICS
Lesson 5: Tangent Lines and the Tangent Function
77
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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 1โ4 (4 minutes)
In this sequence of exercises, students recall the connections between the trigonometric functions and the geometry of
the unit circle.
Exercises 1โ12
The circle shown to the right is a unit circle, and the length of ๐ซ๏ฟฝฬ๏ฟฝ is ๐
๐
radians.
1. Which segment in the diagram has length ๐ฌ๐ข๐ง (๐ ๐
)?
The sine of ๐
๐ is the vertical component of point ๐จ, so ๐จ๐ฉฬ ฬ ฬ ฬ has
length ๐ฌ๐ข๐ง (๐ ๐
).
2. Which segment in the diagram has length ๐๐จ๐ฌ (๐ ๐
)?
The cosine of ๐
๐ is the horizontal component of point ๐จ, so ๐ถ๐ฉฬ ฬ ฬ ฬ ฬ
has length ๐๐จ๐ฌ (๐ ๐
).
3. Which segment in the diagram has length ๐ญ๐๐ง (๐ ๐
)?
Since โณ ๐ถ๐จ๐ฉ is similar to โณ ๐ถ๐ช๐ซ, ๐ช๐ซ
๐=
๐จ๐ฉ
๐ถ๐ฉ=
๐ฌ๐ข๐ง(๐ ๐
)
๐๐จ๐ฌ(๐ ๐
) . Thus, ๐ช๐ซฬ ฬ ฬ ฬ has
length ๐ญ๐๐ง (๐ ๐
).
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 ๐ฌ๐๐(๐ ๐
)?
Since โณ ๐ถ๐จ๐ฉ is similar to โณ ๐ถ๐ช๐ซ, ๐ถ๐ซ
๐ถ๐ฉ=
๐ถ๐ช
๐ถ๐จ. Since ๐ถ๐จ = ๐ถ๐ซ = ๐, we have ๐ถ๐ช =
๐๐ถ๐ฉ
=๐
๐๐จ๐ฌ(๐ ๐
). Thus, ๐ถ๐ชฬ ฬ ฬ ฬ has
length ๐ฌ๐๐(๐ ๐
).
Discussion (7 minutes): Connections Between Geometry and Trigonometry
Do you recall why the length of ๐ถ๐ทฬ ฬ ฬ ฬ is called the tangent of ๐
3? And do you recall why the length of ๐๐ถฬ ฬ ฬ ฬ is called
the secant of ๐
3? (Hint: Look at how lines containing ๐ถ๐ทฬ ฬ ฬ ฬ and ๐๐ถฬ ฬ ฬ ฬ are related to the circle.)
The line containing ๐ถ๐ทฬ ฬ ฬ ฬ is tangent to the circle because it intersects the circle once, so it makes sense to
refer to ๐ถ๐ทฬ ฬ ฬ ฬ as a tangent segment and to refer to the length ๐ถ๐ท as the tangent of ๐
3.
The line containing ๐๐ถฬ ฬ ฬ ฬ is a secant line because it intersects the circle twice, so it makes sense to refer to
๐๐ถฬ ฬ ฬ ฬ as a secant segment and to refer to the length ๐๐ถ as the secant of ๐
3.
How can you be sure that the line containing ๐ถ๐ทฬ ฬ ฬ ฬ is, in fact, a tangent line?
We see that ๐ถ๐ท โก is perpendicular to segment ๐๐ทฬ ฬ ฬ ฬ at point ๐ท, and so ๐ถ๐ท โก must be tangent to the circle.
Explain how this diagram can be used to show that tan (๐
3) =
sin(๐3)
cos(๐3)
.
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
3)
sin(๐
3)
=1
cos(๐
3). If we multiply both sides by sin (
๐3),
we obtain the desired result.
MP.7
NYS COMMON CORE MATHEMATICS CURRICULUM M4 Lesson 5 PRECALCULUS AND ADVANCED TOPICS
Lesson 5: Tangent Lines and the Tangent Function
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Explain how this diagram can be used to show that sin2 (๐3) + cos2 (
๐3) = 1.
If we apply the Pythagorean theorem to โณ ๐ด๐๐ต, we find that (๐ด๐ต)2 + (๐๐ต)2 = (๐๐ด)2, so sin2 (๐3) +
cos2 (๐3) = 1.
Which trigonometric identity can be established by applying the Pythagorean theorem to โณ ๐ถ๐๐ท?
tan2 (๐3) + 12 = sec2 (
๐3)
What is the scale factor that relates the sides of โณ ๐ด๐๐ต to those of โณ ๐ถ๐๐ท?
Point ๐ต is the midpoint of ๐๐ทฬ ฬ ฬ ฬ , so the sides of โณ ๐ถ๐๐ท are twice the length of the corresponding sides of
โณ ๐ด๐๐ต.
Explain how the diagram above can be used to show that tan (๐3) = โ3.
Since the length of ๐ท๏ฟฝฬ๏ฟฝ is ๐
3 radians, it follows that โณ ๐ถ๐๐ธ is equilateral, and so its sides must each be 2
units long. Applying the Pythagorean theorem to โณ ๐ถ๐๐ท, we find 12 + ๐ถ๐ท2 = 22, which means that
๐ถ๐ท2 = 3, and so ๐ถ๐ท = โ3. Thus, tan (๐3) = โ3.
This diagram contains a host of information about trigonometry. But can we take this diagram even further?
Letโs 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!
NYS COMMON CORE MATHEMATICS CURRICULUM M4 Lesson 5 PRECALCULUS AND ADVANCED TOPICS
Lesson 5: Tangent Lines and the Tangent Function
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Exploratory Challenge 1 (7 minutes): Constructing Tangents via Paper-Folding
Earlier we observed that ๐ถ๐ท โก is tangent to the circle. Can you visualize another line through point ๐ถ that is also
tangent to the circle? Try to draw a second tangent line through point ๐ถ, and label its point of intersection
with the circle as point ๐น.
First, letโs examine this diagram through the lens of transformations. Can you see how to map ๐ถ๐ท โก onto ๐ถ๐น โก ?
It appears that you could map ๐ถ๐ท โก onto ๐ถ๐น โก by reflecting it across ๐ถ๐ โก .
Letโs get some hands-on experience with this by performing the reflection in the most natural way
imaginableโby 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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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 ๐ท. Use your straightedge to draw the line through ๐ and ๐ท,
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 ๐ท? Think about this for a moment.
We want to create a line that is perpendicular to ๐๐ท โก at point ๐ท, so we fold the paper in such a way that
the crease is on point ๐ท and ๐๐ท โก maps to itself.
Now choose a point on the crease, and label it ๐ถ. Try to create a second tangent line through point ๐ถ. Can you
see how to do this?
We just need to fold the paper so that the crease is on points ๐ถ and ๐. We can see through the back of
the paper where point ๐ท is, and this is where we mark a new point ๐น, which is the other point of
tangency. In other words, ๐ถ๐น โก is also tangent to the circle.
Do you think you could construct an argument that ๐ถ๐น โก 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 ๐ถ๐ท โก is tangent to the circle,
we know that โ ๐ถ๐ท๐ is a right angle. And since the distance from ๐ to ๐ท is preserved by the reflection,
it follows that point ๐น, which is the image of ๐ท, is also located on the circle because ๐๐น = ๐๐ท, both of
which are radii of the circle. This means that ๐น is on the circle and that โ ๐ถ๐น๐ is a right angle, which
proves that ๐ถ๐น โก is indeed tangent to the circle.
NYS COMMON CORE MATHEMATICS CURRICULUM M4 Lesson 5 PRECALCULUS AND ADVANCED TOPICS
Lesson 5: Tangent Lines and the Tangent Function
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Exploratory Challenge 2 (5 minutes): Constructing Tangents Using a Compass
Letโs 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 ๐๐ท โก at point ๐ท and choose a point ๐ถ on
this perpendicular line as before. Next, we use a compass to draw a circle around point ๐ถ that contains
point ๐ท. This circle will intersect the original circle at a point ๐น, 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 ๐๐น = ๐๐ท. Also, each point on
the circle around point ๐ถ is the same distance from point ๐ถ, so ๐ถ๐น = ๐ถ๐ท. Since ๐ถ๐ = ๐ถ๐,
โณ ๐๐ถ๐ท โ โณ ๐๐ถ๐น by the SSS criterion for triangle congruence. Since โ ๐ถ๐ท๐ is a right angle, it follows by
CPCTC that โ ๐ถ๐น๐ is also a right angle. This proves that ๐ถ๐น โก is indeed tangent to the circle.
Exploratory Challenge 3 (7 minutes): Constructing the Tangents from an External Point
Now letโs 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
NYS COMMON CORE MATHEMATICS CURRICULUM M4 Lesson 5 PRECALCULUS AND ADVANCED TOPICS
Lesson 5: Tangent Lines and the Tangent Function
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This is a tough problem, isnโt it? Letโs 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 โ ๐ถ๐น๐ and
โ ๐ถ๐ท๐ are right angles.
Perhaps we can use this to our advantage. Letโs consider the entire locus of points ๐น such that โ ๐ถ๐น๐ is a right
angle. Can you describe this locus? Can you construct it? This is a challenge in its own right. Letโs 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 ๐ด and ๐ต. Now take a second
piece of paper, and line up one edge on ๐ด and the adjacent edge on ๐ต. Mark the point where the corner of the
paper is, then shift the corner to a new spot, keeping the edges against ๐ด and ๐ต. 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 ๐ด๐ตฬ ฬ ฬ ฬ .
Now construct this circle using your compass.
We just need to construct the midpoint of ๐ด๐ตฬ 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 ๐๐ถฬ ฬ ฬ ฬ , we create two points ๐น and ๐ท with two important properties.
First, they are both on the circle around ๐. Second, they are on the circle with diameter ๐๐ถฬ ฬ ฬ ฬ , which
means that โ ๐๐น๐ถ and โ ๐๐ท๐ถ are right angles, just as we require for tangent lines.
D
F
O
C
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 5โ7 (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:
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 ๐ฉ.
Since point ๐ซ is on circle ๐ช, we know that โ ๐ฉ๐ซ๐จ is a right angle. Since point ๐ซ is on circle ๐ฉ and โ ๐ฉ๐ซ๐จ is a right
angle, it follows that ๐จ๐ซ โก is tangent to circle ๐ฉ. In the same way, we can show that ๐จ๐ฌ โก is tangent to circle ๐ฉ.
7. Use a compass to construct a line that is tangent to the circle below at point ๐ญ. Then choose a point ๐ฎ on the
tangent line, and construct another tangent to the circle through ๐ฎ.
Sample solution:
F
O
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 8โ12 (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 ๐ซ๏ฟฝฬ๏ฟฝ is ๐
๐ radians.
Which trigonometric function corresponds to the length of ๐ฌ๐ญฬ ฬ ฬ ฬ ?
The length of ๐ฌ๐ญฬ ฬ ฬ ฬ represents the cotangent of ๐
๐.
9. Which trigonometric function corresponds to the length of ๐ถ๐ญฬ ฬ ฬ ฬ ?
The length of ๐ถ๐ญฬ ฬ ฬ ฬ represents the cosecant of ๐
๐.
10. Which trigonometric identity gives the relationship between the lengths of the sides of โณ ๐ถ๐ฌ๐ญ?
๐๐จ๐ญ๐ (๐
๐) + ๐๐ = ๐๐ฌ๐๐ (
๐
๐)
11. Which trigonometric identities give the relationships between the corresponding sides of โณ ๐ถ๐ฌ๐ญ and โณ ๐ถ๐ฎ๐จ?
We have ๐
๐ฌ๐ข๐ง(๐
๐)
=๐๐จ๐ญ(
๐
๐)
๐๐จ๐ฌ(๐
๐)
and ๐
๐ฌ๐ข๐ง(๐
๐)
=๐๐ฌ๐(
๐
๐)
๐.
12. What is the value of ๐๐ฌ๐ (๐ ๐
)? What is the value of ๐๐จ๐ญ (๐ ๐
)? Use the Pythagorean theorem to support your
answers.
Let ๐ = ๐ฌ๐ญ. Then we have ๐๐ + ๐๐ = (๐๐)๐ = ๐๐๐, which means ๐๐๐ = ๐; therefore, ๐ = โ๐๐
. So ๐๐จ๐ญ (๐ ๐
) = โ๐๐
and ๐๐ฌ๐ (๐ ๐
) = ๐โ๐๐
.
NYS COMMON CORE MATHEMATICS CURRICULUM M4 Lesson 5 PRECALCULUS AND ADVANCED TOPICS
Lesson 5: Tangent Lines and the Tangent Function
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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.
Letโs say the center of the circle is point ๐ด and the external point is ๐ต. First, we need to construct the
midpoint of the segment joining ๐ด and ๐ตโcall this point ๐ถ. Next, we draw a circle around point ๐ถ that
goes through ๐ด and ๐ต. The points where this circle intersects the first circle are the points of tangency,
so we just connect these points to point ๐ต to form the tangents.
Exit Ticket (4 minutes)
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Lesson 5: Tangent Lines and the Tangent Function
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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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Lesson 5: Tangent Lines and the Tangent Function
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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 ๐ซ and ๐ฌ are on circle ๐ช, โ ๐ฉ๐ซ๐จ and โ ๐ฉ๐ฌ๐จ are right angles. Thus, ๐จ๐ซ โก and ๐จ๐ฌ โก are tangent to circle ๐ฉ.
Problem Set Sample Solutions
1. Prove Thalesโ theorem: If ๐จ, ๐ฉ, and ๐ท are points on a circle where ๐จ๐ฉฬ ฬ ฬ ฬ is a diameter of the circle, then โ ๐จ๐ท๐ฉ is a
right angle.
Since ๐ถ๐จ = ๐ถ๐ท = ๐ถ๐ฉ, โณ ๐ถ๐ท๐จ and โณ ๐ถ๐ท๐ฉ are isosceles triangles. Therefore,
๐โ ๐ถ๐จ๐ท = ๐โ ๐ถ๐ท๐จ, and ๐โ ๐ถ๐ท๐ฉ = ๐โ ๐ถ๐ฉ๐ท.
Let ๐โ ๐ถ๐ท๐จ = ๐ถ and ๐โ ๐ถ๐ท๐ฉ = ๐ท. The sum of three internal angles of โณ ๐จ๐ท๐ฉ
equals ๐๐๐ยฐ.
Therefore, ๐ถ + (๐ถ + ๐ท) + ๐ท = ๐๐๐ยฐ, so ๐๐ถ + ๐๐ท = ๐๐๐ยฐ, and ๐ถ + ๐ท = ๐๐ยฐ. Since
๐โ ๐จ๐ท๐ฉ = ๐ถ + ๐ท, we have ๐โ ๐จ๐ท๐ฉ = ๐๐ยฐ, so โ ๐จ๐ท๐ฉ is a right angle.
NYS COMMON CORE MATHEMATICS CURRICULUM M4 Lesson 5 PRECALCULUS AND ADVANCED TOPICS
Lesson 5: Tangent Lines and the Tangent Function
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2. Prove the converse of Thalesโ theorem: If ๐จ๐ฉฬ ฬ ฬ ฬ is a diameter of a circle and ๐ท is a point so that โ ๐จ๐ท๐ฉ is a right angle,
then ๐ท lies on the circle for which ๐จ๐ฉฬ ฬ ฬ ฬ is a diameter.
Construct the right triangle, โณ ๐จ๐ท๐ฉ.
Construct the line ๐ that is parallel to ๐ท๐ฉฬ ฬ ฬ ฬ through point ๐จ.
Construct the line ๐ that is parallel to ๐จ๐ทฬ ฬ ฬ ฬ through point ๐ฉ.
Let ๐ช be the intersection of lines ๐ and ๐.
The quadrilateral ๐จ๐ช๐ฉ๐ท forms a parallelogram by construction.
By the properties of parallelograms, the adjacent angles are supplementary. Since โ ๐จ๐ท๐ฉ is a right angle, it follows
that angles โ ๐ช๐จ๐ท, โ ๐ฉ๐ช๐จ, and โ ๐ท๐ฉ๐ช are also right angles. Therefore, the quadrilateral ๐จ๐ช๐ฉ๐ท is a rectangle.
Let ๐ถ be the intersection of the diagonals ๐จ๐ฉฬ ฬ ฬ ฬ and ๐ช๐ทฬ ฬ ฬ ฬ . Then, by the properties of parallelograms, point ๐ถ is the
midpoint of ๐จ๐ฉฬ ฬ ฬ ฬ and ๐ช๐ทฬ ฬ ฬ ฬ , so ๐ถ๐จ = ๐ถ๐ฉ = ๐ถ๐ช = ๐ถ๐ท. Therefore, ๐ถ is the center of the circumscribing circle, and the
hypotenuse of โณ ๐จ๐ท๐ฉ, ๐จ๐ฉฬ ฬ ฬ ฬ , is a diameter of the circle.
3. Construct the tangent lines from point ๐ท to the circle given below.
Mark any three points ๐จ, ๐ฉ, and ๐ช on the circle, and construct perpendicular bisectors of ๐จ๐ฉฬ ฬ ฬ ฬ and ๐ฉ๐ชฬ ฬ ฬ ฬ .
Let ๐ถ be the intersection of the two perpendicular bisectors.
NYS COMMON CORE MATHEMATICS CURRICULUM M4 Lesson 5 PRECALCULUS AND ADVANCED TOPICS
Lesson 5: Tangent Lines and the Tangent Function
91
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Construct the midpoint ๐ฏ of ๐ถ๐ทฬ ฬ ฬ ฬ .
Construct a circle with center ๐ฏ and radius ๐ถ๐ฏ.
The circle centered at ๐ฏ will intersect the original circle ๐ถ at points ๐จ and ๐ฉ.
Construct two tangent lines ๐ท๐จ โก and ๐ท๐ฉ โก .
NYS COMMON CORE MATHEMATICS CURRICULUM M4 Lesson 5 PRECALCULUS AND ADVANCED TOPICS
Lesson 5: Tangent Lines and the Tangent Function
92
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This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
4. Prove that if segments from a point ๐ท are tangent to a circle at points ๐จ and ๐ฉ, then ๐ท๐จฬ ฬ ฬ ฬ โ ๐ท๐ฉฬ ฬ ฬ ฬ .
Let ๐ท be a point outside of a circle with center ๐ถ, and let ๐จ and ๐ฉ be points on the circle so that ๐ท๐จฬ ฬ ฬ ฬ and ๐ท๐ฉฬ ฬ ฬ ฬ are
tangent to the circle. Then, ๐ถ๐จ = ๐ถ๐ฉ, ๐ถ๐ท = ๐ถ๐ท, and ๐โ ๐ถ๐จ๐ท = ๐โ ๐ถ๐ฉ๐ท = ๐๐ยฐ, so โณ ๐ท๐จ๐ถ โ โณ ๐ท๐ฉ๐ถ by the
Hypotenuse Leg congruence criterion. Therefore, ๐ท๐จฬ ฬ ฬ ฬ โ ๐ท๐ฉฬ ฬ ฬ ฬ because corresponding parts of congruent triangles are
congruent.
5. Given points ๐จ, ๐ฉ, and ๐ช so that ๐จ๐ฉ = ๐จ๐ช, construct a circle so that ๐จ๐ฉฬ ฬ ฬ ฬ is tangent to the circle at ๐ฉ and ๐จ๐ชฬ ฬ ฬ ฬ is
tangent to the circle at ๐ช.
Construct a perpendicular bisector of ๐จ๐ฉฬ ฬ ฬ ฬ .
Construct a perpendicular bisector of ๐จ๐ชฬ ฬ ฬ ฬ .
The perpendicular bisectors will intersect at point ๐ฏ.
Construct a line through points ๐จ and ๐ฏ.
Construct a circle with center ๐ฏ and radius ๐ฏ๐จฬ ฬ ฬ ฬ ฬ .
The circle centered at ๐ฏ will intersect ๐ฏ๐จ โก at ๐ฐ.
Construct a circle centered at ๐ฐ with radius ๐ฐ๐ฉฬ ฬ ฬ ฬ .