Please explain the exterior angle bisector theorem!!!

## Exterior Angle Bisector Theorem

Exterior angle bisector theorem : The external bisector of an angle of a triangle divides the opposite side externally in the ratio of the sides containing the angle.**Given : **A ΔABC, in which AD is the bisector of the exterior ∠A and intersects BC

produced in D.**Prove that : **BD / CD = AB / AC **Construction : **Draw CE || DA meeting AB in E.

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) |

**Examples **

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.**Given : **AB = 10 cm, AC = 6 cm and BC = 12 cm

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

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72 = 10x – 6x

72 = 4x

x = 72/4

x = 18

CE = 18 cm

PLZ...... GIVE THUMPS UP..

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