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Ch 8 Summary . . . and why SSA and AAA are NOT Δ Congruence Methods of Proof!

Advanced Geometry. Ch 8 Summary . . . and why SSA and AAA are NOT Δ Congruence Methods of Proof!. Why SSA Won’t Work!. Begin with a RIGID line segment (For example: AB = 3 cm) . Create a FIXED angle measure with another ray sharing endpoint A. We have:. AB = 3 cm. B. SIDE.

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Ch 8 Summary . . . and why SSA and AAA are NOT Δ Congruence Methods of Proof!

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  1. Advanced Geometry Ch 8 Summary . . . and why SSA and AAA are NOT ΔCongruence Methods of Proof!

  2. Why SSA Won’t Work! Begin with a RIGID line segment (For example: AB = 3 cm) Create a FIXEDangle measure with another ray sharing endpoint A. We have: AB = 3 cm B SIDE m∡A = 50⁰ ANGLE 3 cm 50⁰ A

  3. Why SSA Won’t Work! Next, let B be the center of circle B with radius “r” B 3 cm r r 50⁰ A C D

  4. Why SSA Won’t Work! Now, since they are radii of the SAME circle, B 3 cm 50⁰ A C D we have established congruent segments, BC and BD.

  5. Why SSA Won’t Work! SIDE BC  BD (radii of ⊙B) AB  AB (reflexive property) SIDE ∡A ∡A (reflexive property) ANGLE B 3 cm 50⁰ A C D Is ΔABCΔABD ?

  6. Why SSA Won’t Work! S BC  BD (radii of ⊙B) AB  AB (reflexive property) S ∡A ∡A (reflexive property) A B 3 cm 50⁰ ∡C is OBTUSE! A C ΔABC

  7. Why SSA Won’t Work! S BC  BD (radii of ⊙B) AB  AB (reflexive property) S ∡A ∡A (reflexive property) A B 3 cm 50⁰ ∡D is ACUTE! A D ΔABD

  8. Why SSA Won’t Work! S BC  BD (radii of ⊙B) AB  AB (reflexive property) S ∡A ∡A (reflexive property) A ↶this ∡B ≇ that ∡B ↷ B B 3 cm 3 cm 50⁰ 50⁰ OBTUSE ∡ ACUTE ∡ A C A D ↻ AC ≇ AD ↺ Is ΔABCΔABD by SSA?

  9. Why SSA Won’t Work! S BC  BD (radii of ⊙B) AB  AB (reflexive property) S ∡A ∡A (reflexive property) A B B 3 cm 3 cm 50⁰ 50⁰ ACUTE ∡ OBTUSE ∡ A C A D Is ΔABCΔABD by SSA? NO! TWO Different TRIANGLES are possible with SSA! TWO Different TRIANGLES!

  10. Why AAA Won’t Work! Begin with ΔABC B 70⁰ 60⁰ 50⁰ A C

  11. Why AAA Won’t Work! Extend two sides of ΔABC B 70⁰ 60⁰ 50⁰ A C

  12. Why AAA Won’t Work! Draw line DE || to BC D Now we have: ∡A  ∡A . . . Reflexive Property B ∡B  ∡D. . . || lines  corr ∡’s  70⁰ 70⁰ ∡C  ∡E. . . || lines  corr ∡’s  60⁰ 50⁰ 50⁰ A C E

  13. Why AAA Won’t Work! ∡A ∡A Is ΔABC ΔADE by AAA ? ∡B ∡D D ∡C ∡E 70⁰ B B 70⁰ 70⁰ 60⁰ 60⁰ 60⁰ 50⁰ 50⁰ 50⁰ A C C E A

  14. Why AAA Won’t Work! ∡A ∡A A With AAA we have proven . . . A ∡B ∡D  means same SHAPE & SIZE D A ∡C ∡E ~ means same SHAPE 70⁰ B ~ 70⁰ = 60⁰ 60⁰ 50⁰ 50⁰ A C E A ΔABC IS SIMILAR TOΔADE !

  15. Theorems from Ch 7 that can easily be related to these concepts! Theorem 50: The sum of the measures of the three angles of a triangle is 180. Theorem 52: A segment joining the midpoints of two sides of a triangle is parallel to and half the length of the third side. Theorem 53: If two angles of one triangle are congruent to two angles of a second triangle, then the third angles are congruent. (NO CHOICE)

  16. Theorems from Ch 7 that can easily be related to these concepts! Theorem 50: The sum of the measures of the three angles of a triangle is 180. You’ve known this one for a long time! . . . Right?

  17. Theorems from Ch 7 that can easily be related to these concepts! Theorem 52: A segment joining the midpoints of two sides of a triangle is parallel to and half the length of the third side. In 8.5 you will learn a new theorem that supports our conclusion as drawn. For now, how could the AAA example easily be adjusted to model Theorem 52?

  18. Theorems from Ch 7 that can easily be related to these concepts! Theorem 53: If two angles of one triangle are congruent to two angles of a second triangle, then the third angles are congruent. (NO CHOICE) In the AAA example, do you think you really needed the measures of all three corresponding angles to prove similarity? Also, how do we know that corresponding sides are proportional?

  19. Chapter 8 Topics Preview 8.1 Ratio and Proportion 8.2 Similarity 8.3 Methods of Proving Triangles Similar 8.4 Congruences and Proportions in Similar Triangles 8.5 Three Theorems Involving Proportions

  20. Chapter 8.1 Theorems 8.1 Ratio and Proportion Definition:Ratio - a quotient of two numbers Definition:Proportion - an equation stating that two or more ratios are equal. Definition:Geometric Mean or Mean Proportional- When the means of a proportion are equal, this is how you refer to either of the means between the extremes. Thm 59: In a proportion, the product of the means is equal to the product of the extremes. a.k.a. [Means-Products Theorem] Thm 60: If the product of a pair of nonzero numbers is equal to the product of another pair of nonzero numbers, then either pair of numbers can be made the extremes, and the other pair the means, of a proportion. a.k.a. [Means-Extremes Ratio Theorem]

  21. Chapter 8.1 Theorems Remember! When solving for ALL values of x that make the equation true, you MUST include BOTH positive and negative values since the variable is squared! 4 8.1 Ratio and Proportion Definition:Ratio - a quotient of two numbers 7 Example: “A rise of 4 for every run of seven” Definition:Proportion - an equation stating that two or more ratios are equal. 4 12 Tripled  = 7 21 Tripled  Definition:Geometric Mean or Mean Proportional- When the means of a proportion are equal, this is how you refer to either of the means between the extremes. x2 = 144 4 x -12 12 If4 : x = x : 36, then = x = ±12 36 -12 12 x

  22. Chapter 8.1 Theorems 8.1 Ratio and Proportion Definition: EXTREMES– the 1st and 4th terms in a proportion Definition: MEANS– the 2nd and 3rd terms in a proportion 1st : 2nd = 3rd : 4th 1st 3rd MEAN EXTREME = Extreme MEAN 2nd EXTREME 4th Extreme MEANS Thm 59: In a proportion, the product of the means is equal to the product of the extremes. a.k.a. [Means-Products Theorem] 12 4 = (7)(12) = (4)(21) 84 = 84 √ 7 21

  23. Chapter 8.1 Theorems 8.1 Ratio and Proportion Thm 60: If the product of a pair of nonzero numbers is equal to the product of another pair of nonzero numbers, then either pair of numbers can be made the extremes, and the other pair the means, of a proportion. a.k.a. [Means-Extremes Ratio Theorem] Extremes MEANS 2 x 6 4 x 3 = 2 3 MEAN Extreme = E : M = M : E 4 6 MEAN Extreme 2 : 4 = 3 : 6 12 = 12 √ The product of the means equals the product of the extremes!

  24. Chapter 8.1 Theorems 8.1 Ratio and Proportion Thm 60: If the product of a pair of nonzero numbers is equal to the product of another pair of nonzero numbers, then either pair of numbers can be made the extremes, and the other pair the means, of a proportion. a.k.a. [Means-Extremes Ratio Theorem] Extremes MEANS 4 x 3 2 x 6 = 4 6 MEAN Extreme = E : M = M : E 2 3 MEAN Extreme 4 : 2 = 6 : 3 12 = 12 √ The product of the means equals the product of the extremes!

  25. Chapter 8.1 Theorems 8.1 Ratio and Proportion Thm 60: If the product of a pair of nonzero numbers is equal to the product of another pair of nonzero numbers, then either pair of numbers can be made the extremes, and the other pair the means, of a proportion. a.k.a. [Means-Extremes Ratio Theorem] MEANS Extremes 2 x 6 3 x 4 = 2 4 MEAN Extreme = E : M = M : E 6 MEAN 3 Extreme 2 : 3 = 4 : 6 12 = 12 √ The product of the means equals the product of the extremes!

  26. Chapter 8.1 Theorems 8.1 Ratio and Proportion Thm 60: If the product of a pair of nonzero numbers is equal to the product of another pair of nonzero numbers, then either pair of numbers can be made the extremes, and the other pair the means, of a proportion. a.k.a. [Means-Extremes Ratio Theorem] Again, if a : b = c : d, and we let a = 2, b = 4, c = 3, and d = 6 a c Extreme MEAN 9 6 3 +6 2 + 4 b d Extreme MEAN = = = a + b c + d 6 4 THEN . . . d b 36 = 36 √ The product of the means equals the product of the extremes!

  27. 8.1 Two Types of MEANS An ARITHMETIC MEAN is when you are finding the AVERAGE of two or more numbers A GEOMETRIC MEAN is when you are finding VALUES that lie between two extremes. EXAMPLE: Find the ARITHMETIC MEANand the GEOMETRIC MEANbetween 3 and 27. ARITHMETIC MEAN: GEOMETRIC MEAN: 3 x 3 + 27 = x 27 2 x2 = 81 30 = 15 = x = 2

  28. 8.1 Two Types of MEANS An ARITHMETIC MEAN is when you are finding the AVERAGE of two or more numbers EXAMPLE: Find the ARITHMETIC MEANand the GEOMETRIC MEANbetween 3 and 27. ARITHMETIC MEAN: 12 to the right 3 + 27 12 to the left 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 30 Middle! = 15 AVERAGE 2

  29. 8.1 Two Types of MEANS A GEOMETRIC MEAN is when you are finding VALUES that lie between two extremes. EXAMPLE: Find the ARITHMETIC MEANand the GEOMETRIC MEANbetween 3 and 27. GEOMETRIC MEAN: When x = - 9 When x = 9 3 x x 9 3 x 3 - 9 x 27 x 9 27 27 x2 = 81 = = = = - 9 = x - 1 1 1 1 x = 3 3 - 3 3 x = CHECK √ CHECK √

  30. Chapter 8.2 Theorems 8.2 Similarity Definition:Reduction - the opposite of an enlargement Definition:Dilation - an enlargement _6__4__2_ 15 10 5 = = 6 4 2 5 15 10 Definition:Similar Polygons - are polygons in which: 1) The RATIOS of the measures of corresponding SIDES are equal. 2) Corresponding ANGLES are CONGRUENT.

  31. Chapter 8.2 Theorems 8.2 Similarity Thm 61: The ratio of the perimeters of two similar polygons equals the ratio of any pair of corresponding sides. Perimeter A = 5(5cm) Perimeter B = 5(15cm) B A 25 cm 5 P = 75cm = 75 cm 15 5 cm 1 3 P = 25cm 15 cm A : B = 1 : 3 …either way!

  32. Chapter 8.3 Theorems 8.3 Methods of Proving Triangles Similar Triangle Similarity POSTULATE: If there exists a correspondence between the vertices of two triangles such that: E 70⁰ B 70⁰ 60⁰ 60⁰ 50⁰ 50⁰ C A D F Postulate:The THREE ANGLES of one triangle are congruent to the corresponding angles of the other triangle, then the triangles are similar. [AAA]

  33. Chapter 8.3 Theorems 8.3 Methods of Proving Triangles Similar Remember, “No Choice!” Triangle Similarity Theorem: If there exists a correspondence between the vertices of two triangles such that: E 70⁰ B 70⁰ 60⁰ 60⁰ C A D F Thm 62: TWO ANGLES of one triangle are congruent to the corresponding angles of the other, then the triangles are similar. [AA]

  34. Chapter 8.3 Theorems 8.3 Methods of Proving Triangles Similar Triangle Similarity Theorem: If there exists a correspondence between the vertices of two triangles such that: E _5_ 10 _6_ 12 _7_ 14 B = = _1_ 2 12 10 5 6 C A D F 14 7 Thm 63: The ratios of the measures of corresponding sides are EQUAL, then the triangles are similar. [SSS~]

  35. Chapter 8.3 Theorems 8.3 Methods of Proving Triangles Similar Triangle Similarity Theorem: If there exists a correspondence between the vertices of two triangles such that: E B _1_ 2 _7_ 14 _6_ 12 = = 12 in 6 in C 50⁰ 50⁰ A D F 7 in 14 in Thm 63: the ratios of the measures of TWO PAIRS of corresponding SIDES are equal, and the INCLUDED ANGLES ARE CONGRUENT, then the triangles are similar. [SAS~]

  36. Chapter 8.4 Theorems 8.4 Congruences and Proportions in Similar Triangles Once you know two triangles are similar, you will use the definition of similar polygons to prove that: 1) Corresponding sides are proportional. Because the ratios of corresponding sides are equal! 2) Corresponding angles of the triangles are congruent. Because they have the same measure!

  37. Chapter 8.5 Theorems 8.5 Three Theorems Involving Proportions CAUTION! Look closely for Midline versus Side-Splitter! Thm 65: If a line is parallel to one side of a triangle and intersects the other two sides, it divides those two sides proportionally. [Side-Splitter Theorem] 5 __10___ 10 + 5 __6__ y _5_ 10 _4_ x = = __6__ y _10__ 15 5x = 40 = y 9 x = 8 10y = 90 6 10 y = 9 8 x 4

  38. Chapter 8.5 Theorems 8.5 Three Theorems Involving Proportions Thm 66: If three or more parallel lines are intersected by two transversals, the parallel lines divide the transversals proportionally. l || m || n || p l KP = 36 K F Find: OP 3 KM = 3x = 3(3) = 9 m G M 3x + 2x + 7x = 36 MO = 2x = 2(3) = 6 2 12x = 36 H O n x = 3 OP = 7x 7 OP = 7(3) J P OP = 21 p 21 + 6 + 9 = 36 ! √

  39. Chapter 8.5 Theorems 8.5 Three Theorems Involving Proportions Thm 67: If a ray bisects an angle of a triangle, it divides the opposite side into segments that are proportional to the adjacent sides. [Angle Bisector Theorem] A PA PE EC Adjacent side AC Adjacent side 14 10 10 5 x 14 10x = 70 = = = Segment Segment P C x = 7 E x 5 7 5 10 √ check! 14 7

  40. Ch 8 Assignments 8.1 Pp 329 – 331 (1 – 13; 15; 17; 20 – 22) 8.2 Pp 336 – 337 (1 – 15) 8.3 Pp 341 – 343 (3; 8 – 10; 16, 20) 8.4 Pp 348 – 350 (3, 5, 7, 8; 9 – 12; 14; 16 – 21) 8.5 Pp 354 – 356 (1, 3, 4, 6, 8, 11, 14, 16, 20, 22) Ch 8 Review: Pp 361 – 364 (1 – 15; 18 – 20; 24, 26)

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