triangles

Asked by  | 12th Mar, 2009, 09:53: AM

Expert Answer:

let side BD be 'x'

then, DC=a-x

from the angle bisecting property in ΔABC, we get,

AB/BD=AC/CD,then

c/x=b/a-x

ca-cx=bx

ca=bx+cx

x=ac/b+c

so,BD=ac/b+c;

& DC= a-x = a-(ac/b+c)

        = (ab+ac-ac)/b+c

so, DC = ab/b+c

Answered by  | 29th May, 2009, 11:52: PM

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