Let $Z = AH \cap BC$, $P'= BF \cap CM$ and $Q'= CF \cap BM$. By Brocard on $(AEHD)$ we have that $ BF \perp CM$, hence $ \angle BP'C = \angle FP'C = 90$, so $P'$ lies on $(FZC)$ and on $(BCDE)$. Hence we have that: $\angle DP'C = \angle DBC = 90 - \angle ACB = 90 - \angle ACZ = \angle ZAC = \angle MAD$, so $AMP'D$ is cyclic. From the cyclic quadrilaterals we get that: $ \angle FP'D = 360 - \angle MP'F - \angle MP'D = 360 - \angle MP'B - (180- \angle MAD) = 270 - (180- \angle ZAC) = 90 + \angle ZAC = \angle AZC + \angle ZAC = 180 - \angle ACZ =180 -\angle DCZ= 180 - \angle FHD.$ Hence $P'$ lies on $(DFH)$, i.e $P' \equiv P$ . Analogously we have that $Q' \equiv Q$. Hence, back to our problem, we have proved that $P= BF \cap CM$ and $Q= CF \cap BM$. Hence by power of point on the cyclic quadrilaterals we have that: $ CP \cdot CM = CD \cdot CA = CE \cdot CH$, hence $MPHE$ is cyclic. Analogously $MQHD$ is cyclic and also $EQPD$ is cyclic because we proved that $P,Q$ lie on $(BCDE)$. Now applying radical axis on $(MPHE), (MQHD), (EQPD)$, we get that $PE, QD , MH$ are concurrent and hence $PE, QD , AH$ are concurrent, as needed.

We also let $P' = BF \cap CM$ and $Q' = CF \cap BM$, and also use Brocard on $(AEHD)$ to conclude $\angle BP'C = 90^{\circ}$ and $\angle CQ'B = 90^{\circ}$. Thus $P'$ and $Q'$ are on $(BCDE)$. We know that $\angle FHD = \angle C$ and that: $\angle FP'D = \angle BP'D = 180^{\circ} - \angle BCD = 180^{\circ} - \angle C$ So $P'$ lies in $(DFH)$ and $CM$ and is also inside $ADE$, which means that $P' = P$. Analogously we can prove that $Q'=Q$. We do some more angle chasing: $\angle DQF = \angle DQC = \angle DBC = \angle DAF$ And arrive that $AQFD$ is cyclic. For a similar reason, $APFE$ is also cyclic, therefore the desired concurrency follows by the radical axis theorem.

Let $\Phi$ be the inversion around $\odot(ADHE)$. Since $\Phi(F)$ is the foot from $A$ to $BC$, we get that $\Phi(\odot(EFH)) = \odot(BEH)$, and so $\Phi(Q) = B$. Since $\Phi$ fixes $\odot(BC)$, it follows that $Q\in\odot(BC)$. Similarly, $P\in\odot(BC)$. Then, the result follows from Pascal on $BQDCPE$.

First we observed that $M$ is circumcenter of $(AEHD)$ By brokard’s theorem on $(AEHD)$ implies that $M$ is orthocenter of $\triangle$$BFC$ Claim : $Q$,$F$ and $C$ collinear Proof : Let $Q’ = CF \cap MB$ , $A’$ be feet of altitude from $A$ Since $M$ is orthocenter of $\triangle$$BFC$ then $CQ’ \bot MB$ Then $\square$$Q’EBC$ is cyclic and since $\square$$Q’FA’B$ is cyclic, $$\angle Q’EC = \angle Q‘BC = \angle Q’FM$$Implies $Q’$ lies on circumcircle of $EFH$ Thus $Q’ \equiv Q$ Hence, $Q$,$F$ and $C$ collinear By Claim implies that $CQ \bot MB$ and $BP \bot MC$ Thus $B,E,Q,P,D$ and $C$ lies on the same circle. Hence result desired from consider radical center of $(EQPD)$ , $(EFH)$ and $(DFH)$