# Monica told her mother to let I be the incenter of the inscribed circle of triangle ABC. Prove that AI/ID = (b+c/a)

I is the incenter of the circle , hence it is the meeting point of the bisectors of <A,

We denote sides AB , BC and AC as c, a and b respectively.

To Prove : AI/ID= (b + c)/a

To prove this we use Angle bisector theorem which states that , in a triangle the bisector of an angle divides the opposite side in the ratio of the sides containing it.

In triangle ABD, BI bisects <B,

hence AB/BD= AI/ID

c/ BD = AI/ ID------(1) (by Angle bisector theorem )

In triangle ACD, CI is bisector of <C

AC/CD = AI/ID

b/CD= AI/ID-------(2)

From equations (1) and (2) c/BD = b/CD

CD/BD =b/c

Adding 1 to both sides , CD/BD +1 = b/c +1

(CD+BD)/BD = ( b + c )/c

BC/BD = (b +c)/c

a/BD = (b +c)/c

or c/BD = (b +c)/a, from equation (1 ) we have c/BD= AI/ID

AI/ID = (b +c)/a

Hence proved

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