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]
