Sides of triangle are AB=5 , AC=13 , BC=14
Let the angle bisector of /_A, AD meets BC at D.
By the Angle Bisector Theorem we know,
(BD)/(DC)=(AB)/(AC) , let BD=x ; DC=y :. x/y=5/13
:. x= 5/13*y and x +y=14:. x= 14-y or
14-y = 5/13*y or (1+5/13)y= 14 or 18/13* y =14 or
y=(13*14)/18=(13*7)/9=91/9= 10 1/9 cm
:.x = 14-10 1/9= 3 8/9 cm. Therefore bisector of /_A
divides the opposie side BC(14) into segments
BD=3 8/9 and DC=10 1/9 cms. Similarly segments of side
AB and AC by the bisectors of /_B and /_C can be measured
in above method.
Let the angle bisector of /_C, CE meets AB at E.
let AE=x ; EB=y :. x/y=(AC)/(BC)=13/14
:. x= 13/14*y and x +y=5:. x= 5-y or
5-y = 13/14*y or (1+13/14)y= 5 or 27/14* y =5 or
y=(5*14)/27=70/27= 2 16/27~~2.6 cm .
:.x = 5-2 16/27= 2 11/27= 2.4 cm. Therefore bisector of /_C
divides the opposite side AB(5) into segments
AE=2.4 and EB=2.6 cms. [Ans]
