$\underset{AB}{\rightarrow}$ +$\underset{AC}{\rightarrow}$ = 2$\underset{AM}{\rightarrow}$
=> $\underset{AM}{\rightarrow}$ = $\underset{BM}{\rightarrow}$ - $\underset{BA}{\rightarrow}$
= $\frac{1}{2}$$\underset{BC}{\rightarrow}$ - $\underset{BA}{\rightarrow}$
Ta lại có:
$\underset{BN}{\rightarrow}$ = $\underset{BA}{\rightarrow}$ + $\underset{AN}{\rightarrow}$
= - $\underset{AB}{\rightarrow}$ + x$\underset{AC}{\rightarrow}$
= -$\underset{AB}{\rightarrow}$ + x ($\underset{AB}{\rightarrow}$ + $\underset{BC}{\rightarrow}$)
= (1-x)$\underset{BA}{\rightarrow}$ + x$\underset{BC}{\rightarrow}$
=> $\underset{AM}{\rightarrow}$.$\underset{BN}{\rightarrow}$
= ($\frac{1}{2}$$\underset{BC}{\rightarrow}$ -$\underset{BA}{\rightarrow}$) [(1-x)$\underset{BA}{\rightarrow}$+x$\underset{BC}{\rightarrow}$]
=$\underset{AM}{\rightarrow}$ . $\underset{BN}{\rightarrow}$
=($\frac{1}{2}$x$a^{2}$ - (1-x)$a^{2}$
=> $\underset{AM}{\rightarrow}$.$\underset{BN}{\rightarrow}$ = (($\frac{1}{2}$x)$a^{2}$
Để AM vuông góc với BN thì $\underset{AM}{\rightarrow}$.$\underset{BN}{\rightarrow}$=0
<=> ($\frac{3}{2}$x -1)$a^{2}$=0
x= $\frac{2}{3}$
Vậy với x = $\frac{2}{3}$ thì AM ⊥ BN.