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Objectives

Objectives. Define congruent polygons Prove that two triangles are congruent using SSS, SAS, ASA, and AAS shortcuts. Definition of Congruence. Congruent figures have the same shape and size Congruent polygons have congruent corresponding parts – matching sides and angles.

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Objectives

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  1. Objectives • Define congruent polygons • Prove that two triangles are congruent using SSS, SAS, ASA, and AAS shortcuts

  2. Definition of Congruence • Congruent figures have the same shape and size • Congruent polygons have congruent corresponding parts – matching sides and angles

  3. Proving Two Triangles Congruent • Using definition: all corresponding sides and angles congruent • Using shortcuts: SSS, SAS, ASA, AAS

  4. Side-Side-Side (SSS) Postulate If the three sides of one triangle are congruent to the three sides of another triangle, then the two triangles are congruent. ΔGHF ≅ ΔPQR

  5. Side-Angle-Side (SAS) Postulate If two sides and the included angle of one triangle are congruent to two sides and the included angle of another triangle, then the two triangles are congruent. ΔBCA ≅ΔFDE

  6. Angle-Side-Angle (ASA) Postulate If two angles and the included side of one triangle are congruent to two angles and the included side of another triangle, then the two triangles are congruent. ΔHGB ≅ ΔNKP

  7. Angle-Angle-Side (AAS) Theorem If two angles and a nonincluded side of one triangle are congruent to two angles and the corresponding nonincluded side of another triangle, then the triangles are congruent. ΔCDM ≅ ΔXGT

  8. SSS Example • AB CD and BC  DA (Given) • AC  AC (Reflexive Property of Congruence) • ∆ABC  ∆CDA by SSS. Explain why ∆ABC  ∆CDA.

  9. SAS Example • XZ VZ and WZ  YZ (Given) • ∠XZY  ∠VZW (Vertical angles are  ) • ∆XZY  ∆VZW (SAS) Explain why ∆XZY  ∆VZW

  10. ASA Example Explain why ∆KLN  ∆MNL KL  MN (Given) KL || MN (Given) ∠KLN  ∠MNL (Alternate Interior Angles are  ) LN  LN (Reflexive Property) ∆KLN  ∆MNL (SAS)

  11. AAS Example JL bisects ∠KJM (Given) ∠KJL ∠MJL (Defn. of angle bisector) ∠K ∠M (Given) JL  JL (Reflexive Property) ∆JKL  ∆JML (AAS) Given: JL bisects ∠KJM. Explain why ∆JKL  ∆JML

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