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AC ≅ DF , BC ≅ EF and ∠B ≅ ∠E ( both 90

∴ Δ ABC ≅ Δ DEF by RHS theorem

Statements |
Reasons |

1) AB = GE | 1) Construction |

2) ∠B = ∠FEG | 2) Each 90^{0} |

3) BC = EF | 3) Given |

4) ΔABC ≅ ΔGEF | 4) By SAS postulate |

5) ∠A = ∠G | 5) CPCTC |

6) AC = GF | 6) If the two angles are congruent then angle opposite to them are equal |

7) AC = DF | 7) Given |

8) DF = GF | 8) By transitive property |

9) ∠D = ∠G | 9) Angles opposite to equal sides in ΔDGF are equal |

10) ∠A = ∠D | 10) From (5) and (9) |

11) ∠B = ∠E | 11) Given |

12)∠A+∠B=∠D+∠E | 12) Adding (10) and (11) |

13) ∠C = ∠F | 13) ∠A + ∠B + ∠C = 180 and ∠D + ∠E + ∠F = 180 ^{0} |

14) ΔABC ≅ ΔDEF | 14) By SAS postulate and from (3) (7) and (13) |

1)

Statements |
Reasons |

1) QM = MR | 1) Given |

2) LM = MN | 2) Given |

3)ML ⊥ PQ | 3) Given |

4) ∠QLM = 90^{0} |
4) By definition of perpendicular |

5) MN ⊥ PR | 5) Given |

6) ∠MNR = 90^{0} |
6) By definition of perpendicular |

7) ∠QLM = ∠MNR | 7) Each 90^{0} |

8) ΔQLM ≅ ΔRNM | 8) RHS theorem |

9) ∠Q = ∠R | 9) CPCTC |

10) PR = PQ | 10) If two angles are equal then side opposite to them are also equal |

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