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Parallel Lines and Proportional Parts Chapter 7 – Lesson 4 Parallel Lines and Proportional Parts

Objective: The students will be able to solve problems by using the Parts of Similar Triangles.

What is…? An Altitude? A median? An angle bisector? B A C

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K In the figure, is an altitude of and is an altitude of Find x if and Answer: JI = 28. Example 5-3a

In the figure, ΔABC ~ ΔFGH. Find the value of x. Answer: x = 14

In the figure, ΔLJK ~ ΔSQR. Find the value of x. MK and TR are corresponding medians and LJ and SQ are corresponding sides. JL = 2x and QS = 2(5) or 10. ~Δ have corr. medians proportional to the corr. sides. Substitution 12 ● 10 = 8 ● 2x Cross Products Property 120 = 16x Simplify. 7.5 = x Divide each side by 16. Answer: x = 7.5

In the figure, is an altitude of and is an altitude of Find x if and Answer: 17.5 Example 5-3c

The side opposite the bisected angle is now in the same proportion as the sides of the angle.

Find x. Since the segment is an angle bisector of the triangle, the Angle Bisector Theorem can be used to write a proportion.. Triangle Angle Bisector Theorem 9x = (15)(6) Cross Products Property 9x = 90 Simplify. x = 10 Divide each side by 9. Answer: x = 10

Find n. Answer: 15

Cross Products Property Find PS and SR. ∆  Bisector Theorem Substitution 40x – 80 = 32x + 160 Cross Products Property Subtraction 8x – 80 = 160 8x = 240 Addition Division x = 30 Substitution PS = 30 – 2 SR = 30 + 5 PS = 28 SR = 25 Simplify

by the ∆  Bisector Theorem. Find AC and DC. by the ∆  Bisector Theorem. Substitute in given values. 4y = 4.5y – 9 Cross Products Theorem –0.5y = –9 Simplify. y = 18 Divide both sides by –0.5. Answer: DC = 9 and AC = 16.

Kahoot!

Lesson Summary: Objective: The students will be able to solve problems by using the Parts of Similar Triangles.

Preview of the Next Lesson: Objective: The students will review for Lesson 7-5 to 7-6 test.

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