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Lesson 1 NYS COMMON CORE MATHEMATICS CURRICULUM M5 GEOMETRY Lesson 1: Thalesβ Theorem Student Outcomes ο§ Using observations from a pushing puzzle, students explore the converse of Thalesβ theorem: If β³ π΄π΅πΆ is a right triangle, then π΄, π΅, and πΆ are three distinct points on a circle with a diameter Μ Μ Μ Μ π΄π΅ . ο§ Students prove the statement of Thalesβ theorem: If π΄, π΅, and πΆ are three different points on a circle with a diameter Μ Μ Μ Μ π΄π΅ , then β π΄π΅πΆ is a right angle. Lesson Notes Every lesson in this module is about an overlay of two intersecting lines and a circle. This will be pointed out to students later in the module, but keep this in mind while presenting the lessons. In this lesson, students investigate what some say is the oldest recorded result, with proof, in the history of geometryβ Thalesβ theorem, attributed to Thales of Miletus (c.624βc.546 BCE), about 300 years before Euclid. Beginning with a simple experiment, students explore the converse of Thalesβ theorem. This motivates the statement of Thalesβ theorem, which students then prove using known properties of rectangles from Module 1. Classwork Opening Students explore the converse of Thalesβs theorem with a pushing puzzle. Give each student a sheet of plain white paper, a sheet of colored cardstock, and a colored pen. Provide several minutes for the initial exploration before engaging students in a discussion of their observations and inferences. Scaffolding: Opening Exercise (5 minutes) Opening Exercise a. Mark points π¨ and π© on the sheet of white paper provided by your teacher. b. Take the colored paper provided, and push that paper up between points π¨ and π© on the white sheet. c. Mark on the white paper the location of the corner of the colored paper, using a different color than black. Mark that point πͺ. See the example below. C A Lesson 1: B Thalesβ Theorem This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from GEO-M5-TE-1.3.0-10.2015 ο§ For students with eyehand coordination or visualization problems, model the Opening Exercise as a class, and then provide students with a copy of the work to complete the exploration. ο§ For advanced learners, explain the paper pushing puzzle, and let them come up with a hypothesis on what they are creating and how they can prove it without seeing questions. 10 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 1 M5 GEOMETRY MP.7 & MP.8 d. Do this again, pushing the corner of the colored paper up between the black points but at a different angle. Again, mark the location of the corner. Mark this point π«. e. Do this again and then again, multiple times. Continue to label the points. What curve do the colored points (πͺ, π«, β¦) seem to trace? Discussion (8 minutes) ο§ What curve do the colored points (πΆ, π·, β¦) seem to trace? οΊ ο§ If that is the case, where might the center of that semicircle be? οΊ ο§ The midpoint of the line segment connecting points π΄ and π΅ on the white paper is the center point of the semicircle. What would the radius of this semicircle be? οΊ ο§ They seem to trace a semicircle. The radius is half the distance between points π΄ and π΅ (or the distance between point π΄ and the midpoint of the segment joining points π΄ and π΅). Can we prove that the marked points created by the corner of the colored paper do indeed lie on a circle? What would we need to show? Have students do a 30-second Quick Write, and then share as a whole class. οΊ We need to show that each marked point is the same distance from the midpoint of the line segment connecting the original points π΄ and π΅. Exploratory Challenge (12 minutes) Allow students to come up with suggestions for how to prove that each marked point from the Opening Exercise is the same distance from the midpoint of the line segment connecting the original points π΄ and π΅. Then offer the following approach. Have students draw the right triangle formed by the line segment between the two original points π΄ and π΅ and any one of the colored points (πΆ, π·, β¦) created at the corner of the colored paper. Then, take a copy of the triangle and rotate it 180Λ about the midpoint of Μ Μ Μ Μ π΄π΅ . A sample drawing might be as follows: Lesson 1: Thalesβ Theorem This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from GEO-M5-TE-1.3.0-10.2015 11 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 1 NYS COMMON CORE MATHEMATICS CURRICULUM M5 GEOMETRY Allow students to read the question posed and have a few minutes to think independently and then share thoughts with an elbow partner. Lead students through the questions on the next page. It may be helpful to have students construct the argument outlined in steps (a)β(b) several times for different points on the same diagram. The idea behind the proof is that no matter which colored point is chosen, the distance from that colored point to the midpoint of the segment between points π΄ and π΅ must be the same as the distance from any other colored point to that midpoint. Exploratory Challenge Choose one of the colored points (πͺ, π«, ...) that you marked. Draw the right triangle formed by the line segment connecting the original two points π¨ and π© and that colored point. Take a copy of the triangle, and rotate it 180Λ about Μ Μ Μ Μ . the midpoint of π¨π© Label the acute angles in the original triangle as π and π, and label the corresponding angles in the rotated triangle the same. Todd says π¨πͺπ©πͺβ² is a rectangle. Maryam says π¨πͺπ©πͺβ² is a quadrilateral, but she is not sure it is a rectangle. Todd is right but does not know how to explain himself to Maryam. Can you help him out? a. What composite figure is formed by the two triangles? How would you prove it? A rectangle is formed. We need to show that all four angles measure ππ°. i. What is the sum of the measures of π and π? Why? ππ°; the sum of the measures of the acute angles in any right triangle is ππ°. ii. How do we know that the figure whose vertices are the colored points (πͺ, π«, β¦) and points π¨ and π© is a rectangle? All four angles measure ππ°. The colored points (πͺ, π«, β¦) are constructed as right angles, and the angles at points π¨ and π© measure π + π, which is ππ°. b. Draw the two diagonals of the rectangle. Where is the midpoint of the segment connecting the two original points π¨ and π©? Why? The midpoint of the segment connecting points π¨ and π© is the intersection of the diagonals of the rectangle because the diagonals of a rectangle are congruent and bisect each other. c. Label the intersection of the diagonals as point π·. How does the distance from point π· to a colored point (πͺ, π«, β¦) compare to the distance from π· to points π¨ and π©? The distances from π· to each of the points are equal. d. Choose another colored point, and construct a rectangle using the same process you followed before. Draw the two diagonals of the new rectangle. How do the diagonals of the new and old rectangle compare? How do you know? One diagonal is the same (the one between points π¨ and π©), but the other is different since it is between the new colored point and its image under a rotation. The new diagonals intersect at the same point π· because diagonals of a rectangle intersect at their midpoints, and the midpoint of the segment connecting points π¨ and π© has not changed. The distance from π· to each colored point equals the distance from π· to each original point π¨ and π©. By transitivity, the distance from π· to the first colored point, πͺ, equals the distance from π· to the second colored point, π«. Lesson 1: Thalesβ Theorem This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from GEO-M5-TE-1.3.0-10.2015 12 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 1 NYS COMMON CORE MATHEMATICS CURRICULUM M5 GEOMETRY e. How does your drawing demonstrate that all the colored points you marked do indeed lie on a circle? For any colored point, we can construct a rectangle with that colored point and the two original points, π¨ and π©, as vertices. The diagonals of this rectangle intersect at the same point π· because diagonals intersect at their midpoints, and the midpoint of the diagonal between points π¨ and π© is π·. The distance from π· to that colored point equals the distance from π· to points π¨ and π©. By transitivity, the distance from π· to the first colored point, πͺ, equals the distance from π· to any other colored point. MP.3 & MP.7 By definition, a circle is the set of all points in the plane that are the same distance from a given center point. Therefore, each colored point on the drawing lies on the circle with center π· and a radius equal to half the length of the original line segment joining points π¨ and π©. ο§ Take a few minutes to write down what you have just discovered, and share that with your neighbor. ο§ We have proven the following theorem: THEOREM: Given two points π΄ and π΅, let point π be the midpoint between them. If πΆ is a point such that β π΄πΆπ΅ is right, then π΅π = π΄π = πΆπ. In particular, that means that point πΆ is on a circle with center π and diameter Μ Μ Μ Μ π΄π΅ . ο§ This demonstrates the relationship between right triangles and circles. THEOREM: If β³ π΄π΅πΆ is a right triangle with β πΆ the right angle, then π΄, π΅, and πΆ are three distinct points on a circle with a diameter Μ Μ Μ Μ π΄π΅ . PROOF: If β πΆ is a right angle, and π is the midpoint between points π΄ and π΅, then π΅π = π΄π = πΆπ implies that a circle with center π and radius π΄π contains the points π΄, π΅, and πΆ. ο§ This last theorem is the converse of Thalesβ theorem, which is discussed on the next page in the Example. Review definitions previously encountered by students as stated in Relevant Vocabulary. Relevant Vocabulary CIRCLE: Given a point πΆ in the plane and a number π > 0, the circle with center πΆ and radius π is the set of all points in the plane that are distance π from the point πΆ. RADIUS: May refer either to the line segment joining the center of a circle with any point on that circle (a radius) or to the length of this line segment (the radius). DIAMETER: May refer either to the segment that passes through the center of a circle whose endpoints lie on the circle (a diameter) or to the length of this line segment (the diameter). CHORD: Given a circle πΆ, and let π and π be points on πΆ. Μ Μ Μ Μ ππ is a chord of πΆ. CENTRAL ANGLE: A central angle of a circle is an angle whose vertex is the center of a circle. Lesson 1: Thalesβ Theorem This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from GEO-M5-TE-1.3.0-10.2015 13 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 1 NYS COMMON CORE MATHEMATICS CURRICULUM M5 GEOMETRY Circle Radius Diameter Chord Central Angle β π₯ β π¦ Point out to students that β π₯ and β π¦ are examples of central angles. Example (8 minutes) Share with students that they have just recreated the converse of what some say is the oldest recorded result, with proof, in the history of geometryβThalesβ theorem, attributed to Thales of Miletus (c.624βc.546 BCE), some three centuries before Euclid. See Wikipedia, for example, on why the theorem might be attributed to Thales although it was clearly known before him: http://en.wikipedia.org/wiki/Thales%27_Theorem. Lead students through parts (a)β(b), and then let them struggle with a partner to determine a method to prove Thalesβ theorem. If students are particularly struggling, give them the hint in the scaffolding box. Once students have developed a strategy, lead the class through the remaining parts of this example. Example In the Exploratory Challenge, you proved the converse of a famous theorem in geometry. Thalesβ theorem states the following: If π¨, π©, and πͺ are three distinct points on a circle, and Μ Μ Μ Μ π¨π© is a diameter of the circle, then β π¨πͺπ© is right. Notice that, in the proof in the Exploratory Challenge, you started with a right angle (the corner of the colored paper) and created a circle. With Thalesβ theorem, you must start with the circle and then create a right angle. Prove Thalesβ theorem. a. Draw circle π· with distinct points π¨, π©, and πͺ on the circle and diameter Μ Μ Μ Μ π¨π©. Prove that β π¨πͺπ© is a right angle. Sample image shown to the right. b. c. MP.1 Μ Μ Μ Μ ). What types of triangles are β³ π¨π·πͺ and Draw a third radius (π·πͺ β³ π©π·πͺ? How do you know? They are isosceles triangles. Both sides of each triangle are radii of circle π· and are, therefore, of equal length. Scaffolding: Using the diagram that you just created, develop a strategy to prove Thalesβ theorem. If students are struggling to develop a strategy to prove Thalesβ theorem, give them this hint: Look at each of the angle measures of the triangles, and see if we can prove πβ π¨πͺπ© is ππ°. d. Label the base angles of β³ π¨π·πͺ as π° and the base angles of β³ π©π·πͺ as π°. Express the measure of β π¨πͺπ© in terms of π° and π°. The measure of β π¨πͺπ© is π° + π°. e. How can the previous conclusion be used to prove that β π¨πͺπ© is a right angle? Draw a third radius, and use the result, also known to Thales, to show that the base angles of an isosceles triangle are congruent. ππ + ππ = πππ because the sum of the angle measures in a triangle is πππ°. Then, π + π = ππ, so β π¨πͺπ© is a right angle. Lesson 1: Thalesβ Theorem This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from GEO-M5-TE-1.3.0-10.2015 14 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 1 M5 GEOMETRY Exercises (5 minutes) Allow students to do Exercises individually and then compare answers with a neighbor. Use this as a means of informal assessment, and offer help where needed. Exercises Μ Μ Μ Μ is a diameter of the circle shown. The radius is ππ. π ππ¦, and π¨πͺ = π ππ¦. π¨π© 1. a. Find πβ πͺ. ππ° b. Find π¨π©. ππ ππ¦ c. Find π©πͺ. ππ ππ¦ Μ Μ Μ Μ Μ is a diameter with center π¨. In the circle shown, π©πͺ 2. a. Find πβ π«π¨π©. πππ° b. Find πβ π©π¨π¬. πππ° c. Find πβ π«π¨π¬. ππ° Closing (2 minutes) Give students a few minutes to explain the prompt to their neighbor, and then call the class together and share. Use this time to informally assess understanding and clear up misconceptions. ο§ Explain to your neighbor the relationship that we have just discovered between a right triangle and a circle. Illustrate this with a picture. οΊ If β³ π΄π΅πΆ is a right triangle and the right angle is β πΆ, then π΄, π΅, and πΆ are distinct points on a circle and Μ Μ Μ Μ π΄π΅ is the diameter of the circle. Lesson 1: Thalesβ Theorem This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from GEO-M5-TE-1.3.0-10.2015 15 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 1 NYS COMMON CORE MATHEMATICS CURRICULUM M5 GEOMETRY Lesson Summary Theorems: ο§ Μ Μ Μ Μ , then β π¨πͺπ© is THALESβ THEOREM: If π¨, π©, and πͺ are three different points on a circle with a diameter π¨π© a right angle. ο§ CONVERSE OF THALESβ THEOREM: If β³ π¨π©πͺ is a right triangle with β πͺ the right angle, then π¨, π©, and πͺ are three distinct points on a circle with a diameter Μ Μ Μ Μ π¨π©. ο§ Therefore, given distinct points π¨, π©, and πͺ on a circle, β³ π¨π©πͺ is a right triangle with β πͺ the right angle if and only if Μ Μ Μ Μ π¨π© is a diameter of the circle. ο§ Given two points π¨ and π©, let point π· be the midpoint between them. If πͺ is a point such that β π¨πͺπ© is right, then π©π· = π¨π· = πͺπ·. Relevant Vocabulary ο§ CIRCLE: Given a point πͺ in the plane and a number π > π, the circle with center πͺ and radius π is the set of all points in the plane that are distance π from the point πͺ. ο§ RADIUS: May refer either to the line segment joining the center of a circle with any point on that circle (a radius) or to the length of this line segment (the radius). ο§ ο§ DIAMETER: May refer either to the segment that passes through the center of a circle whose endpoints lie on the circle (a diameter) or to the length of this line segment (the diameter). Μ Μ Μ Μ is called a chord of πͺ. CHORD: Given a circle πͺ, and let π· and πΈ be points on πͺ. π·πΈ ο§ CENTRAL ANGLE: A central angle of a circle is an angle whose vertex is the center of a circle. Exit Ticket (5 minutes) Lesson 1: Thalesβ Theorem This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from GEO-M5-TE-1.3.0-10.2015 16 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 1 NYS COMMON CORE MATHEMATICS CURRICULUM M5 GEOMETRY Name Date Lesson 1: Thalesβ Theorem Exit Ticket Circle π΄ is shown below. 1. Draw two diameters of the circle. 2. Identify the shape defined by the endpoints of the two diameters. 3. Explain why this shape is always the result. Lesson 1: Thalesβ Theorem This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from GEO-M5-TE-1.3.0-10.2015 17 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 1 M5 GEOMETRY Exit Ticket Sample Solutions Circle π¨ is shown below. 1. Draw two diameters of the circle. 2. Identify the shape defined by the endpoints of the two diameters. 3. Explain why this shape is always the result. The shape defined by the endpoints of the two diameters always forms a rectangle. According to Thalesβ theorem, whenever an angle is drawn from the diameter of a circle to a point on its circumference, then the angle formed is a right angle. All four endpoints represent angles drawn from the diameter of the circle to a point on its circumference; therefore, each of the four angles is a right angle. The resulting quadrilateral is, therefore, a rectangle by definition of rectangle. Problem Set Sample Solutions 1. π¨, π©, and πͺ are three points on a circle, and angle π¨π©πͺ is a right angle. What is wrong with the picture below? Explain your reasoning. Draw in three radii (from πΆ to each of the three triangle vertices), and label congruent base angles of each of the three resulting isosceles triangles. See diagram to see angle measures. In the big triangle (β³ π¨π©πͺ), we get ππ + ππ + ππ = πππ. Using the distributive property and division, we obtain π(π + π + π) = πππ, and π + π + π = ππ. But we also have ππ = πβ π© = π + π. Substitution results in π + π + π = π + π, giving π a value of π β π contradiction. Lesson 1: Thalesβ Theorem This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from GEO-M5-TE-1.3.0-10.2015 18 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 1 NYS COMMON CORE MATHEMATICS CURRICULUM M5 GEOMETRY 2. Show that there is something mathematically wrong with the picture below. Μ Μ Μ Μ , πΆπ© Μ Μ Μ Μ Μ , and πΆπͺ Μ Μ Μ Μ ). Label πβ π©π¨πͺ as π° and πβ π©πͺπ¨ as π°. Also label πβ πΆπ¨πͺ as πΛ and πβ πΆπͺπ¨ Draw three radii (πΆπ¨ as π° since β³ π¨πΆπͺ is isosceles (both sides are radii). If πβ π¨π©πͺ is a right angle (as indicated on the drawing), then π° + π° = ππ°. Since β³ π¨πΆπ© is isosceles, πβ π¨π©πΆ = π° + π°. Similarly, πβ πͺπ©πΆ = π° + π°. Now adding the measures of the angles of β³ π¨π©πͺ results in π° + π° + π° + π° + π° + π° = πππ°. Using the distributive property and division, we obtain π° + π° + π° = ππ°. Substitution takes us to π° + π° = π° + π° + π°, which is a contradiction. Therefore, the figure above is mathematically impossible. 3. Μ Μ Μ Μ is the diameter of a circle of radius ππ miles. If π©πͺ = ππ miles, what is π¨πͺ? In the figure below, π¨π© ππ miles 4. Μ Μ Μ Μ is a diameter. In the figure below, πΆ is the center of the circle, and π¨π« a. Find πβ π¨πΆπ©. ππ° b. If πβ π¨πΆπ© βΆ πβ πͺπΆπ« = π βΆ π, what is πβ π©πΆπͺ? ππ° Lesson 1: Thalesβ Theorem This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from GEO-M5-TE-1.3.0-10.2015 19 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 1 M5 GEOMETRY 5. Μ Μ Μ Μ π·πΈ is a diameter of a circle, and π΄ is another point on the circle. The point πΉ lies on β‘π΄πΈ such that πΉπ΄ = π΄πΈ. Show that πβ π·πΉπ΄ = πβ π·πΈπ΄. (Hint: Draw a picture to help you explain your thinking.) Since πΉπ΄ = π΄πΈ (given), πβ πΉπ΄π· = πβ πΈπ΄π· (both are right angles, β πΈπ΄π· by Thalesβ theorem and β πΉπ΄π· by the angle addition postulate), and π΄π· = π΄π· (reflexive property), then β³ π·πΉπ΄ β β³ π·πΈπ΄ by SAS. It follows that β π·πΉπ΄ β β π·πΈπ΄ (corresponding sides of congruent triangles) and that πβ π·πΉπ΄ = πβ π·πΈπ΄ (by definition of congruent angles). 6. Inscribe β³ π¨π©πͺ in a circle of diameter π such that Μ Μ Μ Μ π¨πͺ is a diameter. Explain why: a. π¬π’π§(β π¨) = π©πͺ. Μ Μ Μ Μ is the hypotenuse, and π¨πͺ = π. Since sine is the ratio of the opposite side to the hypotenuse, π¬π’π§(β π¨) π¨πͺ necessarily equals the length of the opposite side, that is, the length of Μ Μ Μ Μ π©πͺ. b. ππ¨π¬(β π¨) = π¨π©. Μ Μ Μ Μ is the hypotenuse, and π¨πͺ = π. Since cosine is the ratio of the adjacent side to the hypotenuse, ππ¨π¬(β π¨) π¨πͺ necessarily equals the length of the adjacent side, that is, the length of Μ Μ Μ Μ π¨π©. Lesson 1: Thalesβ Theorem This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from GEO-M5-TE-1.3.0-10.2015 20 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.