Cevians AA', BB', and CC' are concurrent at the incenter I of ΔABC.

What is the value of

    |AI|      |BI|      |CI|
   ------- + ------- + -------
    |AA'|     |BB'|     |CC'|

in terms of the side lengths a, b, and c of ΔABC?

Can you prove it?

I agree with Jer. The total is 2.

Drop perpendiculars from A and I to BC and denote their lengths by:

ID = r (the in-radius) and AE = h (an altitude of triangle ABC).

Area of triangle = sum of areas of three sub-triangles. Thus:

ah/2 = ar/2 + br/2 + cr/2  which gives  r/h = a/(a + b + c)           (1)

AI/AA’ = (AA’ - IA’)/AA’ = 1 - IA’/AA’

                                    = 1 -  r/h                       using similar triangles

                                    = 1 - a/(a + b + c)         using (1)

                                    = (b + c)/(a + b + c)

Then, by cyclic interchange of a, b and c:

AI/AA’ + BI/BB’ + CI/CC’ = [(b + c) + (c + a) + (a + b)]/(a + b + c)

                                    = 2

Posted by Harry on 2011-08-25 23:59:22