SLG M4_7_3_1 Similarity Theorem Part 1 PDF

Subject Code Mathematics 4 Module Code 7.0 (Triangle Similarity) Lesson Code 7.3.1 (Similarity Theorems Part 1) Time Limit 30 minutes By the end of this lesson, the student will have been able to: 1. state and illustrate the similarity theorems, in parti...

Subject Code Mathematics 4 Module Code 7.0 (Triangle Similarity) Lesson Code 7.3.1 (Similarity Theorems Part 1) Time Limit 30 minutes By the end of this lesson, the student will have been able to: 1. state and illustrate the similarity theorems, in particular – AAA Similarity Theorem – AA Corollary – SAS Similarity Theorem – SSS Similarity Theorem; and 2. solve problems using the similarity theorems. Recall (Time allotment: 3 minutes ) In the past lessons, we have discussed the definition of similar triangles. When are two triangles similar? Two triangles are similar if the corresponding angles are congruent and the corresponding sides are proportional. Just like when we discussed triangles congruence, this would mean that we have to show that all three corresponding pairs of angles are congruent, and all three pairs of corresponding sides are proportional. Let us consider the two triangles 4ABC and 4GHI. If 4ABC ∼ 4GHI, then it follows that ∠A ∼ = ∠G, ∠B ∼ = ∠H, ∠C ∼ AB BC AC = ∠I, and GH = HI = GI. Is there a simpler way of knowing if two triangles are similar? The following theorems are used to prove triangle similarities. The Similarity Theorems (Time allotment: 7 minutes) AAA Similarity Theorem Given a correspondence between two triangles. If corresponding angles are congruent, then the correspondence is a similarity. (Moise & Downs, 1991) Restatement. Given a correspondence between two triangles 4ABC and 4DEF. If ∠A ∼ = ∠D, ∼ ∼ ∠B = ∠E, and ∠C = ∠F, then we say that 4ABC ∼ 4DEF. Proof. We are given that the corresponding pairs of angles are all congruent, we need to establish that AB BC AC the corresponding sides are proportional. This means we need to show that DE = EF = DF. AB BC Let us first show the first part of the equation, i.e., DE = EF. Consider 4ABC and 4DEF below. B E A C D F Figure 1 Mathematics 4 Page 1 of 7  − → − → Let D0 and E 0 be distinct points on BA and BC, respectively, such that D0 B = DE and BF 0 = EF. Refer to the following figures. B E D0 F0 A C D F Figure 2 Therefore, by SAS, we can conclude that 4BD0 F 0 ∼= 4EDF. Hence, ∠BD0 F 0 ∼ = ∠D. Since ∠D ∼ = ∠A, 0 0 ∼ 0 0 ∼ then we say that ∠BD F = ∠A. In Figure 2 above, if ∠BD F = ∠A, what can we conclude? By the Basic Proportionality Theorem, we can say that DAB BC 0 0 0 B = BF 0. But since D B = DE and BF = EF , then we AB BC can conclude that DE = EF. BC AC We can prove the second part of the equation, i.e., EF = DF , by doing the same process. This time, we −→ −→ will consider two points, D and E on CA and CB such that CD0 = FD and CE 0 = EF. Following the 0 0 BC AC same flow of the proof, the goal is to establish that EF = DF. The proof of this part is left as an exercise for the student. By proving the AAA Similarity Theorem, we can now simply say that if we have two triangles for which all pairs of corresponding angles are congruent, then these two triangles are similar. The AAA Similarity Theorem has two corollaries. We will discuss only one of these corollaries. AA Corollary (Time allotment: 3 minutes) AA Corollary Given a correspondence between two triangles. If two pairs of corresponding angles are congruent, then the correspondence is a similarity. (Moise & Downs, 1991) Since this is a corollary of the previous theorem, it means that the proof of this will apply the AAA Similarity Theorem. Again, the proof of AA Corollary is left for you as an exercise. You need to recall the Corollary to the Triangle Sum Theorem. If you already know that two pairs of corresponding angles of two triangles are congruent, what can you conclude about the third pair of corresponding angles? Now this theorem needs only two pairs of corresponding angles of two triangles to be congruent. With the said condition, you can already conclude that the two triangles are similar. Let us now discuss another similarity theorem which is quite similar to one of the congruence postulates that we discussed in the First Quarter. Mathematics 4 Page 2 of 7 But before we discuss this theorem, let us first state a theorem that we will be using in the proof. Converse of Basic Proportionality Theorem If a line intersects two sides of a triangle, and cuts off segments proportional to these two sides, then it is parallel to the third side. (Moise & Downs, 1991) The proof of this theorem can be found at the end of the module as supplementary reading. SAS Similarity Theorem (Time allotment: 7 minutes) SAS Similarity Theorem Given a correspondence between two triangles. If two pairs of corresponding sides are proportional, and their included angles are congruent, then the correspondence is a similarity. (Moise & Downs, 1991) AB BC Restatement. Given 4ABC and 4DEF, with the correspondence ABC ←→ DEF. If DE = EF and ∠B ∼ = ∠E, then 4ABC ∼ 4DEF. Proof. Consider 4ABC and 4DEF below. B E A C D F Figure 3 − → − → Let D0 and F 0 be distinct point on BA and BC, respectively, such that D0 B = DE and BF 0 = EF. B E D0 F0 A C D F Figure 4 Again, by SAS we conclude that that 4D0 BF 0 ∼ = 4DEF. Therefore, DAB BC 0 B = BF 0 , (Why?). By the Converse 0 0 of the Basic Proportionality Theorem, we have D F k AC. Since we have parallel sides, then by the Mathematics 4 Page 3 of 7  ∼ ∠A. By the Reflexive Property, ∠B = Converse of Corresponding Angle Theorem, ∠BD0 F 0 = ∼ ∠B. 0 0 0 0 ∼ Therefore, 4ABC ∼ 4D BF (Why?).But we also know that 4D BF = 4DEF. Hence, it is logical to say that 4ABC ∼ 4DEF. We will now discuss the last of the similarity theorems which is called the SSS Similarity Theorem. SSS Similarity Theorem (Time allotment: 10 minutes) SSS Similarity Theorem Given a correspondence between two triangles. If corresponding sides are proportional, then the correspondence is a similarity. (Moise & Downs, 1991) AB BC AC Restatement. Given 4ABC and 4DEF, with the correspondence ABC ←→ DEF. If DE = EF = DF , then 4ABC ∼ 4DEF. We will again use the same figure (Figure 3) as before. B E A C D F Figure 5 − → − → Let D0 and F 0 be distinct point on BA and BC, respectively, such that D0 B = DE and BF 0 = EF. B E D0 F0 A C D F Figure 6 You are given a two–column proof of this theorem and asked to provide the statements and reasons that are missing. Recall what you have learned in proving triangle congruence (Module 5.0) to be able to complete the proof. Mathematics 4 Page 4 of 7 Proof of SSS Similarity Theorem (Non-graded) Statement Reason AB BC AC 1. DE = EF = DF 1. 2. DB0 = DE and BF 0 = EF 2. 3. 3. Substitution 4. ∠B ∼ = ∠B 4. 5. 5. SAS Similarity Theorem (Steps 3 and 4) AC AB 6. D0 F 0 = D0 B 6. D0 F 0 D0 B 7. AC = AB 7. 0 8. D0 F 0 = AC DABB = AC DE AB 8. 9. 9. MPE (Step 1) 10. D0 F 0 = DF 10. 11. 11. SSS Postulate (Steps 2 and 10) 12. 4ABC ∼ 4DEF 12. Substitution (Steps 5 and 11) We have now discussed the different similarity theorems that would help us in proving that two triangles are similar. In the next session, we will apply these theorems in practice exercises. This two–part lesson will end with a summative assessment. References Albarico, J.M. (2013). THINK Framework. Based on Ramos, E.G. and Apolinario, N. (n.d.). Science LINKS. Quezon City: Rex Bookstore Inc. Moise, E. E., & Downs, F. L., Jr. (1991). Geometry. Addison-Wesley Publishing Company. Prepared by: Mary Gay Antonette G. Magpantay Reviewed by: Jonellyn S. Albano Position: Special Science Teacher III Position: Special Science Teacher V Campus: PSHS-MC Campus: PSHS-IRC c 2020 Philippine Science High School System. All rights reserved. This document may contain proprietary information and may only be released to third parties with approval of management. Document is uncontrolled unless otherwise marked; uncontrolled documents are not subject to update notification. Mathematics 4 Page 5 of 7 Supplementary Reading Proof of Converse of Basic Similarity Theorem Restatement. Given 4ABC. Let D be a point between A and B, and let E be a point between A and C. AB AC If AD = AE , then DE k BC. A D E B C Figure 7 ←→ ← → Proof. Let BC0 be the line parallel to DE through B, which intersects AC at C0. By the Basic Proportionality AB AC0 Theorem, AD = AE. A D E C0 B C Figure 8 AC AC0 AB AC Since we are given that AD = AE , then we will have AE = AE. Therefore AC = AC0. Hence, point C is the 0 same as point C. Thus, DE k BC. Mathematics 4 Page 6 of 7 Answer Key Proof of SSS Similarity Theorem Statement Reason AB BC AC 1. DE = EF = DF 1. Given 2. DB0 = DE and BF 0 = EF 2. Given AB BC 3. D0 B = BF 0 3. Substitution 4. ∠B ∼ = ∠B 4. Reflexive Property 5. 4ABC ∼ 4D”BF 0 5. SAS Similarity Theorem (Steps 3 and 4) AC AB 6. D0 F 0 = D0 B 6. Definition of Similar Triangles D0 F 0 D0 B 7. AC = AB 7. MPE 0 8. D0 F 0 = AC · DABB = AC · DE AB 8. Symmetric Property, Substitution (Steps 2 and 7) 9. DF = AC·DE AB 9. MPE (Step 1) 10. D0 F 0 = DF 10. Substitution (Steps 8 and 9) 11. 4DEF ∼ 4D0 BF” 11. SSS Postulate (Steps 2 and 10) 12. 4ABC ∼ 4DEF 12. Substitution (Steps 5 and 11) Mathematics 4 Page 7 of 7