Pearson Unit 1 Topic 4: Congruent Triangles 4-1: Congruent Figures Pearson Texas Geometry ©2016 Holt Geometry Texas ©2007

TEKS Focus: (6)(C) Apply the definition of congruence, in terms of rigid transformations, to identify congruent figures and their corresponding sides and angles. (1)(B) Use a problem-solving model that incorporates analyzing given information, formulating a plan or strategy, determining a solution, justifying the solution, and evaluating the problem-solving process and the reasonableness of the solution. (1)(F) Analyze mathematical relationships to connect and communicate mathematical ideas. (1)(G) Display, explain, or justify mathematical ideas and arguments using precise mathematical language in written or oral communication.

CONGRUENT POLYGONS Figures that have congruent corresponding sides and corresponding angles. A B C D E F G H

When you write a statement such as ABC  DEF, you are also stating which parts are congruent. Helpful Hint D A F B C E

Example: 1 Given: ∆WYS  ∆MKV Identify all pairs of corresponding congruent parts. Angles: W  M, Y  K, S  V Sides: WY  MK, YS  KV, SW  VM

Example: 2 If polygon LMNP  polygon EFGH, identify all pairs of corresponding congruent parts. Angles: L  E, M  F, N  G, P  H Sides: LM  EF, MN  FG, NP  GH, LP  EH

Example: 3 Example 3 mS = 180 – (62 + 35) = 83 mS = mV = 83

TR = AN 4x – 1 = 2x + 7 2x – 1 = 7 2x = 8 x = 4 TI = AG TI = 3x + 4 Example 4 TR = AN 4x – 1 = 2x + 7 2x – 1 = 7 2x = 8 x = 4 TI = AG TI = 3x + 4 TI = 3(4) + 4 TI = 12 + 4 TI = 16

Example: 5 Given: ∆ABC  ∆DBC. Find the value of x. BCA and BCD are rt. s. Def. of  lines. BCA  BCD Rt.   Thm. mBCA = mBCD Def. of  s Substitute values for mBCA and mBCD. (2x – 16)° = 90° 2x = 106 Add 16 to both sides. x = 53 Divide both sides by 2.

Write this theorem after Example 5:

Example: 6 Given: ∆ABC  ∆DBC. Find mDBC. mABC + mBCA + mA = 180° ∆ Sum Thm. mABC + mBCA + mA = 180° Substitute values for mBCA and mA. mABC + 90 + 49.3 = 180 mABC + 139.3 = 180 Simplify. Subtract 139.3 from both sides. mABC = 40.7 DBC  ABC Corr. s of  ∆s are  . mDBC = mABC Def. of  s. mDBC  40.7° Trans. Prop. of =

Example: 7 Given: ∆ABC  ∆DEF Find the value of x. AB  DE Corr. sides of  ∆s are . AB = DE Def. of  parts. Substitute values for AB and DE. 2x – 2 = 6 2x = 8 Add 2 to both sides. x = 4 Divide both sides by 2.

Example: 8 Given: ∆ABC  ∆DEF Find mF. ∆ Sum Thm. mEFD + mDEF + mFDE = 180° ABC  DEF Corr. s of  ∆ are . mABC = mDEF Def. of  s. mDEF = 53° Transitive Prop. of =. Substitute values for mDEF and mFDE. mEFD + 53 + 90 = 180 mF + 143 = 180 Simplify. mF = 37° Subtract 143 from both sides.

Example: 9 Given: YWX and YWZ are right angles. YW bisects XYZ. W is the midpoint of XZ. XY  YZ. Prove: ∆XYW  ∆ZYW

Statements Reasons 1. YWX and YWZ are rt. s. 1. Given 2. YWX  YWZ 2. Rt.   Thm. 3. YW bisects XYZ 3. Given 4. XYW  ZYW 4. Def. of  bisector 5. W is midpoint of XZ 5. Given 6. XW  ZW 6. Def. of midpoint 7. Reflexive Prop. of  7. YW  YW 8. X  Z 8. Third s Thm. 9. XY  YZ 9. Given 10. ∆XYW  ∆ZYW 10. Def. of  ∆s

Example 10: Given: AD bisects BE. BE bisects AD. AB  DE, A  D Prove: ∆ABC  ∆DEC

Statements Reasons 1. A  D 1. Given 2. BCA  DCE 2. Vertical s are . 3. ABC  DEC 3. Third s Thm. 4. Given 4. AB  DE AB  DE DE AB  DE 5. Given BE bisects AD 5. AD bisects BE, 6. BC  EC, AC  DC 6. Def. of bisector 7. ∆ABC  ∆DEC 7. Def. of  ∆s