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AB ≅ DE , BC ≅ EF and AC ≅ DF

∴ ΔABC ≅ Δ DEF by SSS

1) In triangle ABC, AD is median on BC and AB = AC.

Prove that ∠ABD = ∠ACD

Statements |
Reasons |

1) AB = AC | 1) Given |

2) AD is a median | 2) Given |

3) BD = DC | 3) By definition of median. |

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

5) ΔADC ≅ ΔADB | 5) By SSS postulate |

6) ∠ABD = ∠ACD | 6) CPCTC |

2) ΔABC and ΔDBC are two isosceles triangle on the same base BC and vertices of A and D are on the same side of BC. If AD is extended to intersect BC at P.

Prove that i) ΔABD ≅ ΔACD ii) AP is the perpendicular bisector of BC.

⇒ AB = AC and BD = DC

Statements |
Reasons |

1) AB = AC | 1) Given |

2) BD = CD | 2) Given |

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

4) ΔABD ≅ ΔACD | 4) By SSS postulate |

5) ∠BAP = ∠CAP | 5) By CPCTC ∠BAD = ∠CAD |

6) AP = AP | 6) Reflexive(common side) |

7) ΔBAP ≅ ΔCAP | 7) By SAS postulate |

8) ∠APB = ∠APC | 8) CPCTC |

9) BP = CP | 9) CPCTC |

10) ∠APB + ∠APC = 180 | 10) These two angles are linear pair angles and they are supplementary |

11) 2∠APB = 180 | 11) Addition property |

12) ang;APB = 90^{0} |
12) Division property |

13) AP is the perpendicular bisector of BC | 13) By definition of perpendicular bisector and from (9) and (12) |

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