AB ≅ DE, BC ≅ EF and ∠B ≅ ∠E

∴ ΔABC ≅ Δ DEF by SAS

Statements |
Reasons |

1) AB = AC | 1) Given |

2) AD is a bisector | 2) By construction |

3) ∠BAD = ∠CAD | 3) By definition of angle bisector |

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

5) ΔABD ≅ ΔACD | 5) SAS Postulate |

6) ∠B = ∠C | 6) CPCTC |

1) O is the mid point of AB and CD. Prove that

i) ΔAOC ≅ ΔBOD ii) AC = BD and iii) AC || BD .

Statements |
Reasons |

1) O is the mid point. | 1) Given |

2) AO = OB | 2) By definition of mid point. |

3) ∠AOC = ∠BOD | 3) Vertically opposite angles |

4) CO = OD | 4) By definition of mid point. |

5) ΔAOC ≅ ΔBOD | 5) SAS postulate |

6) AC = BD | 6) CPCTC |

7) ∠CAO = ∠DBO | 7) CPCTC |

8) AC || BD | 8) If alternate interior angles are congruent then the lines are parallel. |

2) If D is the mid point of the hypotenuse AC of a right triangle ABC, prove that BD = ½ AC.

Statements |
Reasons |

1) AD = DC | 1) Given |

2) BD = DE | 2) By construction |

3) ∠ADB = ∠CDE | 3) Vertically opposite angles |

4) ΔADB ≅ ΔCDE | 4) By SAS postulate |

5) EC = AB and ∠CED = ∠ABD | 5) CPCTC |

6) CE || AB | 6) If alternate interior angles are congruent then the lines are parallel |

7) ∠ABC + ∠ECB = 180 | 7) Angles formed on the same side of transveral are supplementary. |

8) 90 + ∠ECB = 180 | 8) Since ∠B = 90 given |

9) ∠ECB = 180 -90 = 90 | 9) By subtraction property |

10) AB = EC | 10) From (5) |

11) BC = CB | 11) Reflexive (Common side) |

12) ∠ABC = ∠ECB | 12) Each 90^{0} |

13) ΔABC ≅ ΔECB | 13) SAS postulate |

14) AC = BE | 14) CPCTC |

15) 1/2AC = 1/2BE ⇒ 1/2AC = BD | 15) Multiply by 1/2 but 1/2BE = BD by mid point definition. |

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