There are three cases :

Statements |
Reasons |

1) AB = DE | 1) According to 1st case |

2) ∠A = ∠D | 2) Given |

3) ∠B = ∠E | 3) Given |

4) ΔABC ≅ ΔDEF | 4) By ASA postulate |

⇒ AB = DE, BC = EF and AC = DF

Consequently, Q coincides with F.

AB BC CA ---- = ------ = ------ DE EF FA |

Since the corresponding angles are equal, we conclude that Δ ABC ~ Δ DEF.

In triangles ABC and DPQ,

Statements |
Reasons |

1) AB = DP | 1) By construction |

2) ∠A = ∠D | 2) Given |

3) AC = DQ | 3) By construction |

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

5) ∠B = ∠DPQ | 5) CPCTC |

6) ∠B = ∠E | 6) Given |

7) ∠E = ∠DPQ | 7) By transitive property ( from above) |

8) PQ || EF | 8) If two corresponding angles are congruent then the lines are parallel |

9) DP/DE = DQ/DF | 9) By basic proportionality theorem |

10) AB/DE = BC/EF | 10) By construction |

11) AB/DE = AC/DF | 11) By substitution property |

12) Δ ABC ~ Δ DEF | 12) By SAS postulate |

Proof for this case is same as above case ( ii ).

• AAA Similarity

• AA Similarity

• SSS Similarity

• SAS Similarity

• Practice on Similarity

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