In the triangle ABC, $MN \parallel BC$, $BO$ and $CO$ are angle bisectors of $\angle MBC$ and $\angle NCB$ respectively. If $AB=12$, $BC=24$ and $AC=18$, then what is the perimeter of $\triangle AMN$ ?
I made an aproximated drawing
I tried with similarity but got nothing there. Maybe i'm not seeing the similar triangles in the correct way. Any hints in this kind of problem?
Thanks.
On
The semi-perimeter is $s = {12+18+24 \over 2} = 27$
The area of $\triangle$ ABC is, by Heron, $\sqrt{27 . 15 . 9 . 3 } = 27 \sqrt{15} $
The inradius is $ r = {Area \over s} = \sqrt{15} $ = the elevation of O.
The elevation of A is $ {2Area\over 24 } = {54\over 24 }\sqrt{15} $
The similarity ratio comes out from the ratio of the two elevations $= {54\over (54-24) }$
The semi-perimeter of $\triangle$ AMN is $= {30 \over 54} 27 = 15 $
The required perimeter is then $30$.
It's better the following way: $$P_{\Delta AMN}=AM+MO+AN+NO=AM+MB+AN+NC=12+18=30.$$