∠B ≅ ∠E , ∠C ≅ ∠F and AC ≅ DF

∴ Δ ABC ≅ Δ DEF by AAS

Statements |
Reasons |

1) ∠B = ∠C | 1) Given |

2) ∠BAD = ∠CAD | 2) By construction |

3) AD = AD | 3) Reflexive (common side) |

4) ΔABD ≅ ΔACD | 4) By angle angle side postulate (AAS) |

5) AB = AC | 5) By CPCTC |

1) If ΔABC is an isosceles triangle with AB = AC. Prove that the perpendiculars from the vertices B and C to their opposite sides are equal.

Statements |
Reasons |

1) AB = AC | 1) Given |

2) ∠ABC = ∠ACB | 2) If two sides are congruent then the angle opposite to them are also congruent |

3) ∠CEB = ∠BDC | 3) Each 90^{0} |

4) BC = BC | 4) Reflexive (common side) |

5) ΔBCE ≅BCD | 5) By AAS postulate |

6) BD = CE | 6) CPCTC |

2) ∠A = ∠C and AB = BC. Prove that ΔABD ≅ ΔCBE.

Statements |
Reasons |

1) ∠A = ∠C | 1) Given |

2) ∠AOE = ∠COD | 2) Vertically opposite angles |

3) ∠A + ∠AOE = ∠C + ∠COD | 3) Add (1) and (2) |

4) 180^{0} - ∠AEO = 180^{0} - ∠CDO |
4) Since ∠A + ∠AOE + ∠AEO = 180 and ∠C + ∠COD + ∠CDO = 180 |

5) ∠AEO = ∠CDO | 5) By subtraction property |

6) ∠AEO + ∠OEB = 180^{0} |
6) Linear pair angles |

7) ∠CDO + ∠ODB = 180^{0} |
7) Linear pair angles |

8) ∠AEO + ∠OEB = ∠CDO + ∠ODB | 8) Transitive property |

9) ∠OEB = ∠ODB | 9) Subtraction property and from(5) |

10) ∠CEB = ∠ADB | 10) Since ∠OEB = ∠CEB and ∠ODB = ∠ADB |

11) AB = BC | 11) Given |

11) ΔABD ≅ ΔCBE | 11) By AAS postulate (from (1),(10)) |

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