Let $2\alpha,2\beta,2\gamma$ be the angles $\angle BAC, \angle CBA, \angle ACB,$ respectively. We first prove that $EA = ED.$ We will do so by showing that $\angle EAD = \angle EDA.$ WLOG assume $AB < AC;$ the case when $AB > AC$ is symmetrical to that of $AB < AC,$ and $AB = AC$ is impossible. Let $X$ be the midpoint of arc $BC$ not containing $A.$ First note that $$\angle EAD = \angle EAX = \angle ACX,$$which is the measure of minor arc $AX.$ But this is just equal to $\angle ACB + \frac{\angle BCX}{2} = \alpha + 2\gamma.$ On the other hand, by sum of angles in $\triangle ABD,$ we know that $\angle EDA = 180^\circ - \alpha - 2\beta.$ Thus it suffices to show that $180^\circ- \alpha - 2\beta = \alpha + 2\gamma,$ which is obvious since $\alpha + \beta + \gamma = 90^\circ.$ Therefore, $AE = ED.$ Consequently, $EM \perp AD.$ This also means that $M$ lies on $(EAH)$ since $\angle EHA = \angle EMA =90^\circ.$ Now let $R$ be an arbitrary point on $r$ closer to $BC$ than $M.$ Then $$\angle RMA = 180^\circ - \angle RMD = 180^\circ - (90^\circ - \angle ADB) = 90^\circ + \angle ADB.$$On the other hand, we have $$\angle MEA = 90^\circ - \angle EAM = 90^\circ - \angle ADE.$$Therefore, $\measuredangle RMA = \measuredangle MEA,$ done.

Because $E$ is the center of the Apollonius Circle of $A$ relative to $BC$ and $AB/AC = DB/DC$, it follows that $EA=ED$, from which we get that $\angle AME = 90^{\circ} = \angle AHE$, thus $AMHE$ is inscribed in a circumference whose center is the midpoint of $AE$; call that center $N$. Now, since $MN$ is a midline of $ADE$, $MN$ is parallel to $BC$ and thus perpendicular to $r$, which gives the desired result.