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Proportions & Similar Triangles

This lesson focuses on the principles of proportionality theorems used in geometry, specifically regarding similar triangles. Students will learn how to calculate segment lengths and solve real-life problems by applying these theorems. Key concepts include the Triangle Proportionality Theorem and its converse, as well as methods to determine segment lengths using cross multiplication. Practical examples are provided to illustrate the application of proportionality in various scenarios, including determining dimensions and using angle bisectors effectively.

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Proportions & Similar Triangles

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  1. Proportions & Similar Triangles

  2. Objectives/Assignments • Use proportionality theorems to calculate segment lengths. • To solve real-life problems, such as determining the dimensions of a piece of land.

  3. Use Proportionality Theorems • In this lesson, you will study four proportionality theorems. Similar triangles are used to prove each theorem.

  4. Triangle Proportionality Theorem If a line parallel to one side of a triangle intersects the other two sides, then it divides the two side proportionally. If TU ║ QS, then RT RU = TQ US

  5. Converse of the Triangle Proportionality Theorem If a line divides two sides of a triangle proportionally, then it is parallel to the third side. RT RU If , then TU ║ QS. = TQ US

  6. Ex. 1: Finding the length of a segment • In the diagram AB ║ ED, BD = 8, DC = 4, and AE = 12. What is the length of EC?

  7. Step: DC EC BD AE 4 EC 8 12 4(12) 8 6 = EC Reason Triangle Proportionality Thm. Substitute Multiply each side by 12. Simplify. = = EC = • So, the length of EC is 6.

  8. Ex. 2: Determining Parallels • Given the diagram, determine whether MN ║ GH. LM 56 8 = = MG 21 3 LN 48 3 = = NH 16 1 8 3 ≠ 3 1 MN is not parallel to GH.

  9. Proportional parts of // lines • If three parallel lines intersect two transversals, then they divide the transversals proportionally. • If r ║ s and s║ t and l and m intersect, r, s, and t, then UW VX = WY XZ

  10. Special segments • If a ray bisects an angle of a triangle, then it divides the opposite side into segments whose lengths are proportional to the lengths of the other two sides. • If CD bisects ACB, then AD CA = DB CB

  11. In the diagram 1  2  3, and PQ = 9, QR = 15, and ST = 11. What is the length of TU? Ex. 3: Using Proportionality Theorems

  12. SOLUTION: Because corresponding angles are congruent, the lines are parallel. PQ ST Parallel lines divide transversals proportionally. = QR TU 9 11 = Substitute 15 TU 9 ● TU = 15 ● 11 Cross Product property 15(11) 55 TU = = Divide each side by 9 and simplify. 9 3 • So, the length of TU is 55/3 or 18 1/3.

  13. In the diagram, CAD  DAB. Use the given side lengths to find the length of DC. Ex. 4: Using the Proportionality Theorem

  14. Since AD is an angle bisector of CAB, you can apply Theorem 8.7. Let x = DC. Then BD = 14 – x. Solution: AB BD = Apply Thm. 8.7 AC DC 9 14-X Substitute. = 15 X

  15. Ex. 4 Continued . . . 9 ● x = 15 (14 – x) 9x = 210 – 15x 24x= 210 x= 8.75 Cross product property Distributive Property Add 15x to each side Divide each side by 24. • So, the length of DC is 8.75 units.

  16. In the diagram KL ║ MN. Find the values of the variables. Finding Segment Lengths

  17. To find the value of x, you can set up a proportion. Solution 9 37.5 - x = Write the proportion Cross product property Distributive property Add 13.5x to each side. Divide each side by 22.5 13.5 x 13.5(37.5 – x) = 9x 506.25 – 13.5x = 9x 506.25 = 22.5 x 22.5 = x • Since KL ║MN, ∆JKL ~ ∆JMN and JK KL = JM MN

  18. To find the value of y, you can set up a proportion. Solution 9 7.5 = Write the proportion Cross product property Divide each side by 9. 13.5 + 9 y 9y = 7.5(22.5) y = 18.75

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