Geometric problem about regular triangle We have a regular triangle $ABC$, where side is $15$, point $N$ belongs to $AB$ such that $AN = 5$. Point $M$ belongs to $AC$ such that $AM = 3$. Prove that $BM$ is perpendicular to $CN$.
I tried to make similar triangles carrying the $BK$ segment, where $K$ is the midpoint of $MC$. So, as $NB:AN = MK:AM$ and angle $A$ is common, $ANM$ and $ABK$ triangles are similar, right? Follows that
$BK \| NM$.
I have multiple versions of solving, e.g. if we prove that $MK=KC=OK$, where $O$ is the intersection point of $BM$ and $CN$, then $MOC$ triangle will become right, or we can show that $COM$ and $BMH$, where $H$ is the base of height from $B$, triangles are similar.
So, I can't conclude what to do.
 A: Here is a solution which arguably can be considered elementary.
Let $\Omega$ be the intersection $BM\cap CN$, and draw the cevian $AL$ through $\Omega$ from $A$, $L$ being its intersection with $BC$.

Then $BL=5$ and $LC=10$, since for this placement of $L$ with $BL:LC=1:2$ we have the relation (Ceva)
$\displaystyle -1=
\frac{NA}{NB}\cdot
\frac{LB}{LC}\cdot
\frac{MC}{MA}$, which is explicitly forgetting about signs (that match)
$\displaystyle -1=
\frac{ 5}{10}\cdot
\frac{ 5}{10}\cdot
\frac{12}{ 3}$.
Draw from $L$ parallels $LM'\|BM$ and $LN'\|CN$ with $M'\in AC$, $N'\in AB$.
Then project $N',A,M'$ on $BC$, obtaining the points $N'',A'',M''$.
Using similarities we can compute all lengths as marked in the picture.
Moreover, $CM''$ takes from $CA''=\frac{15}2$ a proportion, which is the same one as $CM':CA=8:15$, so $CM''=\frac 82=4$. From here $LM''=10-4=6$. Also  $M'M'':AA''= CM':CA=8:15$ and $AA''=\frac{15\sqrt3}2$ is giving $M'M''=\frac{8\sqrt3}2=4\sqrt 3$.
Doing the same on the other side we get
$BN''=\frac{BN'}{BA}BA''=\frac{10/3}{15}\cdot\frac{15}2=\frac 53$, so
$N''L=5-\frac 53=\frac{10}3$. And
$N'N''=\frac{BN'}{BA}AA''=\frac{10/3}{15}\cdot\frac{15\sqrt 3}2=\frac {5\sqrt 3}3$, so
$N''L=5-\frac 53=\frac{10}3$.
It remains to check the similarity $\Delta N''LN'\sim\Delta M''M'L$, the triangles have each a right angle, and we check the proportion:
$$
\frac{N''L}{N''N'}=
\frac{10/3}{5\sqrt 3/3}=
\frac 2{\sqrt 3}=
\frac{4\sqrt 3}{6}=
\frac{M''M'}{M''L}
\ .
$$
From here $90^\circ=\widehat{N'LM'}=\widehat{N\Omega M}$.
$\square$
A: Expanding a little bit con my comment, let $H$ and $K$ be the projections of $M$ and $C$ on $AB$. Let $O = CN \cap BM$ and $P = CK\cap BM$.


*

*Compute $\overline{MH} = \frac{3\sqrt 3}2$, and $\overline{HB} = 15-\frac32 =
    \frac{27}2$.

*Compute $\overline{CK} = \frac{15\sqrt 3}2$, and $\overline{NK} = \frac{15}2-5 = \frac52$.

*Observe that $$\frac{\overline{HB}}{\overline{MH}} = \frac{\overline{CK}}{\overline{NK}} = 3\sqrt3,$$hence $\triangle MHB \sim\triangle CNK$ by SAS criterion. In particular $\angle NCK \cong \angle MBH$.

*By 3. and the fact that $\angle KPB \cong \angle CPO$ (vertical angles) we get that $\triangle COP \sim \triangle PKB$ by AA criterion. Hence the thesis.

