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
<
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
