I found a lot of nice proofs for this problem. The sufficiency is easy (a corollary of Bramagupta theorem) The necessity : Proof 1 From $\angle HMA = \angle HNA,\angle DHB = \angle EHC$ we can show that triangle $DHB$ is similar to triangle $EHC$,from which we conclude that $D,E,B,C$ are concycllc. Then drawing the circumcenter $O$ of triangle $DHB$,from Bramagupta theorem we obtain that $OH$ is perpendicular to $EC$.Combining with $MH$ perpendicular to $EC$ and $O$ lies on the perpendicular bisector of segment $BD$ we obtain that $O$ concoins with $M$ ,which implies $BE \perp CD$. Proof 2 We prove the conclusion by "swifting" the statement.Noting that swifting $BA$ won't change the conclusion,we can move $BA$ to $CB'$ ,but $A'$ change into $C$,then the midpoint of the image of $BD$ changes into $M'$.And from the orthocenter assertion we know that $H$ is also the orthocenter of $M'NC$,from which we conclude that $M'N \perp HC$ ,so by midline we know that $BE \perp CD$.Then easy angle chasing show that $B,C,E,D$ are concyclic. For some reason now I cant post the attachment...

Alternatively, to prove the necessity case, after showing that $D,E,B,C$ are concyclic, one can easily angle chase to find that the $MH, NH$ is the $H$ - symmedian of $\triangle HCE$ and $\triangle HBD$ respectively. If the symmedian is perpendicular to the opposite side, and the triangle is not isoceles, then it must be right-angled.

Hermitianism wrote: Proof 1 From $\angle HMA = \angle HNA,\angle DHB = \angle EHC$ Why are these angles equal? Edit: oops I'm stupid

Hermitianism wrote: I found a lot of nice proofs for this problem. The sufficiency is easy (a corollary of Bramagupta theorem) The necessity : Proof 1 From $\angle HMA = \angle HNA,\angle DHB = \angle EHC$ we can show that triangle $DHB$ is similar to triangle $EHC$,from which we conclude that $D,E,B,C$ are concycllc. Then drawing the circumcenter $O$ of triangle $DHB$,from Bramagupta theorem we obtain that $OH$ is perpendicular to $EC$.Combining with $MH$ perpendicular to $EC$ and $O$ lies on the perpendicular bisector of segment $BD$ we obtain that $O$ concoins with $M$ ,which implies $BE \perp CD$. Proof 2 We prove the conclusion by "swifting" the statement.Noting that swifting $BA$ won't change the conclusion,we can move $BA$ to $CB'$ ,but $A'$ change into $C$,then the midpoint of the image of $BD$ changes into $M'$.And from the orthocenter assertion we know that $H$ is also the orthocenter of $M'NC$,from which we conclude that $M'N \perp HC$ ,so by midline we know that $BE \perp CD$.Then easy angle chasing show that $B,C,E,D$ are concyclic. For some reason now I cant post the attachment... How do you prove that the triangles are similiar

by creating Phantom point(sorry,I forgot to wrote that)

The orthocenter proving part is just some obvious angle chasing so we'll just prove the other part. Note that $\angle HME = \angle 90 - \angle A = \angle HND$ and $\angle EHB = \angle DHC$ and $HM$ and $HN$ are bisectors of $EB$ and $DC$ so $HEB$ and $HDC$ are similar so $\angle HEB = \angle HDC$ so $BCDE$ is cyclic. Let $NH$ and $MH$ meet $AM$ and $AN$ at $S$ and $K$. Note that $\angle NHC = \angle EHS = \angle 90 - \angle BEH = \angle 90 - \angle CDH = \angle DHK = \angle MHB$ so $MH$ and $NH$ are symmedians in $BEH$ and $CDH$ and Note that these symmedians are perpendicular to $EB$ and $DC$ so $EHB$ and $DHC$ are right angled triangles so $BE \perp CD$.