Statements |
Reasons |

1) CE || DA | 1) By construction |

2) ∠1 = ∠3 | 2) Alternate interior angle |

3) ∠2 = ∠4 | 3) Corresponding angle (CE ||DA and BK is a transversal |

4) AD is a bisector of ∠A | 4) Given |

5) ∠1 = ∠2 | 5) Definition of angle bisector |

6) ∠3 = ∠4 | 6) Transitivity (from 2 and 4) |

7) AE = AC | 7) If angles are equal then side opposite to them are also equal |

8) BD / CD = BA/EA | 8) By Basic proportionality theorem(EC ||AD) |

9) BD /CD = AB/AE | 9) BA = AB and EA = AE |

10) BD /CD = AB /AC | 10) AE = EC and from(7) |

1) In the given figure, AE is the bisector of the exterior ∠CAD meeting BC produced in E. If AB = 10 cm, AC = 6 cm and BC = 12 cm, find CE.

By exterior angle bisector theorem

BE / CE = AB / AC

(12 + x) / x = 10 / 6

6( 12 + x ) = 10 x [ by cross multiplication]

72 + 6x = 10x

<

72 = 10x – 6x

72 = 4x

x = 72/4

x = 18

CE = 18 cm

2) The bisector of interior ∠A of ΔABC meets BC in D, and the bisector of exterior ∠A meets BC produced in E. Prove that BD / BE = CD / CE.

Statements |
Reasons |

1) AD is the internal bisector of ∠A | 1) Given |

2) AB / AC = BD / DC | 2) By Internal angle bisector theorem |

3) AE is an external bisector of ∠A | 3) Given |

4) AB/AC = BE / CE | 4) By External angle bisector theorem |

5) BD / DC = BE / CE | 5) From (2) and (4) |

6) BD/BE = CD/CE | 6) By alternendo ( a/b=c/d ⇒ a/c = b/d) |

• Similarity in Geometry

• Properties of similar triangles

• Basic Proportionality Theorem(Thales theorem)

• Converse of Basic Proportionality Theorem

• Interior Angle Bisector Theorem

• Exterior Angle Bisector Theorem

• Proofs on Basic Proportionality

• Criteria of Similarity of Triangles

• Geometric Mean of Similar Triangles

• Areas of Two Similar Triangles

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