"an angle bisector of an exterior angle of ? abc intersects the line ab at point d. prove that ad: bd=ac: bc."

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

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.

Given : In ΔABC, AD and AE are respectively the bisectors of the interior and exterior angles at A.

Prove that : BD/BE = CD/CE