11. No, need  MKJ   MKL 12. Yes, by Alt Int Angles  SRT   UTR and  STR   URT; RT  RT (reflex) so ΔRST  ΔTUR by ASA 13.  A   D Given  C 

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11. No, need  MKJ   MKL 12. Yes, by Alt Int Angles  SRT   UTR and  STR   URT; RT  RT (reflex) so ΔRST  ΔTUR by ASA 13.  A   D Given  C   F AAS 14. No need to know  K and  H are rt  s 15. Yes, BE  CE and AE  DE;  A   D (rt  s  thm) so ΔRST  ΔTUR by HL 20. Proof B incorrect. Corr. sides are not in correct order 24. Since we know 2 sides and included angle, we could use SAS. Since the Δs are rt  s, we could use HL. 26. A 27. J 28. C 38. AB = 6, BC = °

Warm Up 1. If ∆ABC  ∆DEF, then A  ? and BC  ?. 2. What is the distance between (3, 4) and (–1, 5)? 3. If 1  2, why is a||b? 4. List methods used to prove two triangles congruent. D D EF 17 Converse of Alternate Interior Angles Theorem SSS, SAS, ASA, AAS, HL

CPCTC is an abbreviation for the phrase “Corresponding Parts of Congruent Triangles are Congruent.” It can be used as a justification in a proof AFTER you have proven two triangles congruent. SSS, SAS, ASA, AAS, and HL use corresponding parts to prove triangles congruent. CPCTC uses congruent triangles to prove corresponding parts congruent. Remember!

Example 1 A landscape architect sets up the triangles shown in the figure to find the distance JK across a pond. What is JK? One angle pair is congruent, because they are vertical angles. Two pairs of sides are congruent, because their lengths are equal. Therefore the two triangles are congruent by SAS. By CPCTC, the third side pair is congruent, so JK = 41 ft.

Example 2: Prove: XYW  ZYW Given: YW bisects XZ, XY  YZ. StatementsReasons 1. YW bisects XZ1. 2. XW  ZW XY  YZ YW  YW ΔXYW  ΔZYW  XYW   ZYW 6. Given Def of bisects Given Reflexive SSS CPCTC

Work backward when planning a proof. To show that ED || GF, look for a pair of angles that are congruent. Then look for triangles that contain these angles. Helpful Hint

Example 3 Prove: KL || MN Given: J is the midpoint of KM and NL. StatementsReasons 1. J is the midpoint of KM and NL. 1. Given 2. KJ  MJ, NJ  LJ 5. LKJ  NMJ 4. ∆KJL  ∆MJN 3. KJL  MJN 6. KL || MN 5. CPCTC 6. Conv. Of Alt. Int. s Thm. 4. SAS 3. Vert. s Thm. 2. Def. of mdpt.

Use the Distance Formula to find the lengths of the sides of each triangle. Example 4: Given: D(–5, –5), E(–3, –1), F(–2, –3), G(–2, 1), H(0, 5), and I(1, 3) Prove: DEF  GHI

So DE  GH, EF  HI, and DF  GI. Therefore ∆DEF  ∆GHI by SSS, and DEF  GHI by CPCTC.