Angle $B$ of an acute triangle ABC measures $60^o$.In its interior is located the point
P such that PA + PB + PC is the minimum; let $I$, and $l_2$, are the incenters of the triangles $ABP$ and $BPC$ repectively. Through $I$, e $I_2$ we trace parallels to $BP$ which intersect at $M$ and $N$ to $AB$ and $BC$ respectively; if $BM =a$. Find $BN$ (Answer:$a$.)
My progress:
Corrected drawing with the auxiliary instruction to generate Fermat's point as per article provided by ACB
$\triangle CJN \sim \triangle CPB \implies\frac{CN}{BC} = \frac{CJ}{CP}=\frac{JN}{BM}\\
\triangle AMH \sim \triangle ABP \implies\frac{AM}{AB} = \frac{AH}{AP}=\frac{MH}{BP}$
Th. Thales:$\frac{CN}{BN} = \frac{CJ}{JP} \implies BN = \frac{CN.JP}{CJ}$
I don't know if there is a property that gives the sum of the smallest distance from an interior point of a triangle to the vertices
Best Answer
$I$ and $J$ are incenters of $\triangle APB$ and $\triangle BPC$ respectively.
As $MI$ is parallel to $BP$ and $I$ is incenter,
$\displaystyle \frac{BM}{MX} = \frac{PI}{IX} = \frac{BP}{BX}$
and similarly, $ ~ \displaystyle \frac{BN}{NY} = \frac{PJ}{JY} = \frac{BP}{BY}$
So if we can show that $BX = BY$, that would show $BM = BN$ and we are done.
By some angle chasing, we can show that $\triangle PBX \sim \triangle PCY$ and I leave this as an exercise for you.
Then, $ \displaystyle \frac{BX}{BP} = \frac{CY}{CP}$
But given $PY$ is angle bisector of $\angle BPC$,
$ \displaystyle \frac{CY}{CP} = \frac{BY}{BP}$
$ \therefore BX = BY ~ $ and that completes the proof that $BM = BN$.