Question

prove that the external bisector of any angle of a triangle divides the side opposite to the angle in the ratio of the other two sides

Answer 1

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)

Answer 2

Answer:

Step-by-step explanation:

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