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

Statements |
Reasons |

1) AB AC ---- = ------ DE DF |
1) Given |

2) DP DQ ---- = ------ DE DF |
2) As AB = DP and AC = DQ. By substitution |

3) PQ || EF | 3) By converse of basic proportionality theorem |

4) ∠DPQ = ∠E and ∠DQP = ∠F | 4) Corresponding angles |

5) ΔDPQ ~ ΔDEF | 5) By AA similarity |

6) DP PQ ---- = ------ DE EF |
6) By definition of similar triangles |

7) AB PQ ---- = ------ DE EF |
7) As DP = AB , by substitution |

8) PQ BC ---- = ------ EF EF |
8) From (1) (6) and (7) |

9) PQ = BC | 9) From (8) |

10) ΔABC ≅ ΔDPQ | 10) By S-S-S postulate |

11) ΔABC ~ ΔDEF | 11) From (5) and (10) |

Statements |
Reasons |

1) AB = 4; AC = 2;CB = 6; DE = 2; DF = 1; FE = 3 | 1) Given |

2) AB BC CA ---- = ------ = ------ DE EF FD |
2) Property of proportion |

3) ΔABC ~ ΔDEF | 3) By S-S-S-similarity |

1) In ΔPQR ~ ΔXYZ, PQ = 5cm, QR = 4cm and PR= 6m ,find XY, YZ and XZ.

2) In ΔABC, D and E are any points on AB and AC respectively, such that DE||BC. If AD=x cm, DB=x-2 cm, AE=x-1 cm, then find the value of x.

• AAA Similarity

• AA Similarity

• SSS Similarity

• SAS Similarity

• Practice on Similarity

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