A homothety at $A$, with ratio $2$ tells us that it is sufficient to prove that if the midpoint of $BC$ is $M$, then $EM$ is parallel to $AN$ and its length equals half of $AN$. Take a homothety at $L$ with ratio $2$. Then we need to prove that if the reflection of $L$ over $M$ is $X$, then $AKXN$ is a parallelogram. The spiral similarity at $B$ sending $KA$ to $XC$ implies that there is another spiral similarity at $B$ sending $KX$ to $AC$, so $KXB \sim ACB$. Simple angle chasing reveals, then, that $KX \parallel AN$. Similarly, we have $AK \parallel NX$. So we're done.