2.  TEKS Focus:  (6)(B) Prove two triangles are congruent by applying the Side-Angle-Side, Angle-Side-Angle, Side-Side- Side, Angle-Angle-Side, and Hypotenuse-Leg congruence conditions.  (1)(F) Analyze mathematical relationships to connect and communicate mathematical ideas.  (1)(A) Apply mathematics to problems arising in everyday life, society, and the workplace.  (1)(D) Communicate mathematical ideas, reasoning, and their implications using multiple representations, including symbols, diagrams, graphs, and language as appropriate.  (1)(G) Display, explain, or justify mathematical ideas and arguments using precise mathematical language in written or oral communication.

3. An included side is the common side of two consecutive angles in a polygon. The following postulate uses the idea of an included side.

4. 4-3

5. Determine if you can use ASA to prove the triangles congruent. Explain. Two congruent angle pairs are given, but the congruent side (by the Reflexive Property of congruence) is NOT the included side for the given congruent angles. Therefore ASA cannot be used to prove the triangles congruent.

6. You can use the Third Angles Theorem to prove another congruence relationship based on ASA. This theorem is Angle-Angle-Side (AAS). 4-2

7. Determine if you can use ASA to prove  NKL   LMN. Explain. By the Alternate Interior Angles Theorem.  KLN   MNL. NL  LN by the Reflexive Property. No other congruence relationships can be determined, so ASA cannot be applied.

8. Determine if you can use AAS to prove the triangles congruent with the given information. Explain. Given: LJ bisects  KLM,  K   M  K   M, and  KLJ   MLJ by the definition of bisectors. LJ  LJ by the reflexive property of congruence. Therefore we can use AAS to prove the triangles congruent.

9. ΔHOG  ΔATC by ASA. (Note: FN is not the included side in ΔNFI.

10. There is a given pair of congruent angles; a pair of vertical angles congruent, and PR and SR are the non-included sides. ΔPAR  ΔSIR by AAS.

11.  ABC   DCB by Alt. Int.  s Thm. BC  BC by Reflexive Property of  If x = 30, then  ACB   CBD. ΔABC  ΔDCB by ASA. 3x – 10 = 2x + 20 x – 10 = 20 x = 30

12. STATEMENTSREASONS 1. 2. 3. 4.

13. STATEMENTSREASONS 1. 2. 3. 4. 5.