How Do You Relate These Points Now?

We claim that $\angle EDN=\angle MDG$ .

Connect $GD$ which interesects $\odot ABC$ at $L,G$ . Since $GM,GD$ are isogonals wrt $\angle BGC$ because $GD$ is a symmedian of $GBC$ , therefore $AL,AF$ are isogonals wrt $\angle BAC$ , which means $L\in AO$ and $DG\perp AG$ . Moreover lines $DF,DG$ are symmetric about $DM$ .

We now show $DGNO\sim DMEF$ by individually proving $GON\sim MFE, DGO\sim DMF$ .

The first is trivial by property of circumcenter. For the second, we have $\angle ODG=\angle FDM$ by symmetry mentioned above. Since $OG=OL$ , thus $\angle OGD=\angle OLG=\angle AFD=\angle DMF$ and the similarity is established.

From the similarity we have $\angle EDM=\angle NDG\implies \angle EDN=\angle ODG$ , which is what we desired to show. By some angle chasing or the fact that $OL^2=OC^2=OM\cdot OD$ , we can obtain $\angle ODG=\angle OLM$ . Hence:

$\angle EDN+\angle HMB=\angle OLM+90-\angle MHE =90-(\angle LHE-\angle ALH)=90-\angle HAO=90-|\angle.B-\angle C|=\boxed {70}$