6. In an isosceles $\triangle ABC$, $AD$ is the altitude to the base $BC$, $BE$ is the altitude to $AC$, $AD$ and $BE$ intersect at $H$, $EF \perp BC$ intersects $BC$ at $F$, $M$ is a point on the extension of $AD$ such that $DM = EF$, and $N$ is the midpoint of $AH$. Let $MN^2 = b$, $BN^2 = m$, $BM^2 = n$. The relationship between $b$, $m$, and $n$ is ( ).
(A) $bm+n$
(D) $b \neq m+n$
6. B
6. Without losing generality, we can take an isosceles right $\triangle ABC$ (as shown in the figure). Then points $A, E, H, N$ coincide, and points $D, F$ coincide: It is easy to see that $\triangle BMA$ is the same as $\triangle NBM$, which are isosceles right triangles, $\angle NBM=90^{\circ}$, hence $b=m+n$.