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Class Greeting
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Parallel Lines and Proportional Parts
Chapter 7 – Lesson 5 Parallel Lines and Proportional Parts
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Objective: The students will solve problems using Parallel Lines and Proportional Parts.
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In and Find SU. S 8x = 36 x = 4.5 Example 4-1a
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Substitute in the given values.
Find PN. Use the Triangle Proportionality Theorem. Substitute in the given values. 2PN = 15 Cross Products Prop. PN = 7.5 Divide both sides by 2.
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Example 2: Verifying Segments are Parallel
Verify that Converse of the Triangle Proportionality Theorem
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AC = 36 cm, and BC = 27 cm. Verify that Converse of the Triangle Proportionality Theorem
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In and AZ = 32. Determine whether Explain.
Answer: No; the segments are not in proportion since Example 4-2c
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A midsegment of a triangle is a segment that joins the midpoints of two sides of the triangle. Every triangle has three midsegments, which form the midsegment triangle.
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Example 1: Examining Midsegments in the Coordinate Plane
The vertices of ∆XYZ are X(–1, 8), Y(9, 2), and Z(3, –4). M and N are the midpoints of XZ and YZ. Show that and Step 1 Find the coordinates of M and N. (3, –4) (9, 2) (–1, 8)
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Example 1 Continued Step 2 Compare the slopes of MN and XY. X(–1, 8), Y(9, 2), M(1, 2) and N(6, -1). Since the slopes are the same,
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Example 1 Continued Step 3 Compare the lengths of MN and XY.
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Your Turn The vertices of ΔRST are R(–7, 0), S(–3, 6), and T(9, 2). M is the midpoint of RT, and N is the midpoint of ST. Show that and Step 1 Find the coordinates of M and N.
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Your Turn Step 2 Compare the slopes of MN and RS. Since the slopes are equal
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Your Turn Step 3 Compare the lengths of MN and RS. The length of MN is half the length of RS.
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The relationship shown in Example 1 is true for the three midsegments of every triangle.
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Example 2A: Using the Triangle Midsegment Theorem
Find . BD ∆ Midsegment Theorem Substitution BD = 8.5 Simplify.
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Example 2B: Using the Triangle Midsegment Theorem
Find mCBD ∆ Midsegment Theorem mCBD = mBDF Alternate Interior s Theorem mCBD = 26° Substitution
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Your Turn Find . JL ∆ Midsegment Theorem 36 Substitution 2(36) = JL Multiplication 72 = JL Simplify.
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Given: E, D, and F are midpoints of respectively.
Prove: C Proof: Statements Reasons 1. Given E, D, and F are midpoints of respectively. 1. 2. Triangle Midsegment Theorem 2. 3. Alternate Interior Angles Theorem 3. Example 6-1b
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4. Corresponding Angles Postulate 4.
Continued Proof: Statements Reasons 4. Corresponding Angles Postulate 4. 5. Transitive Property 5. 6. AA Similarity 6. B E D F A C Example 6-1c
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Given: Z and X are the midpoints of respectively.
Prove: Given: Z and X are the midpoints of respectively. is formed by connecting the midpoints of the sides and of Prove that w Example 6-1d
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2. Triangle Midsegment Theorem
Continued Proof: Statements Reasons 1. Given 2. Triangle Midsegment Theorem 3. Corresponding Angles Postulate Z and X are the midpoints of respectively. 4. Reflexive Property 5. AA Similarity 1. 2. 3. 4. 5. Example 6-1e
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Given: M is the midpoint of JK; N is the midpoint of KL; and P is the midpoint of JL.
Prove: ∆JKL ~ ∆NPM Statements Reasons 1. M is the midpoint of JK; N is the midpoint of KL; and 1. Given P is the midpoint of JL. 2. 2. ∆ Midsegs. Thm 3. 3. Division Prop. of =. 4. ∆JKL ~ ∆NPM 4. SSS ~
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In the figure, Larch, Maple, and Nuthatch Streets are all parallel
In the figure, Larch, Maple, and Nuthatch Streets are all parallel. The figure shows the distances in city blocks that the streets are apart. Find x. Answer: x = 32 Example 4-4a
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In the figure, Davis, Broad, and Main Streets are all parallel
In the figure, Davis, Broad, and Main Streets are all parallel. The figure shows the distances in city blocks that the streets are apart. Find x. Answer: x = 5 Example 4-4c
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Kahoot!
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Lesson Summary: Objective: The students will solve problems using Parallel Lines and Proportional Parts. .
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Preview of the Next Lesson:
Objective: The students will be able to use Proportional Relationships, Dilation and Similarity in the Coordinate Plane.
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Stand Up Please
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