Let $H$ be the orthocenter of $ABC$. $\angle BAH = 90^{\circ} - B = \angle MPB = \angle QPC$ $\angle ABH = 90^{\circ} - A = \angle AQM = \angle PQC$ $\triangle ABH \sim \triangle PQC$ Since $M$ and $N$ are the midpoints of $AB$ and $PQ$, $\triangle MBH \sim \triangle NQC$ $\angle QNC = \angle BMH$ $\angle QNC + \angle QMH = \angle BMH + \angle QMH = 90^{\circ}$ $MH \perp DC$ Since $CH \perp AB$ and $A, B, M, D$ are on the same line, $CH \perp MD$ $H$ must be the orthocenter of $\triangle DCM$. $\blacksquare$

Let $Y$ and $Z$ be the feet of the $A$ and $B-$altitude on $BC$ and $AC$ respectively. Let $R=(CYZ)\cap (ABC)$. It is well known that $M,H,R$ are collinear. Let $D'=CR\cap (BC)$. So clearly $H$ is the orthocenter of $\triangle CD'M$ because $CH\perp D'M, MH\perp D'C$. We will prove that $\triangle CMD'$ and $\triangle CMD$ are same or equivalently $D'\equiv D$. Notice that $-1=(QP;N\infty)\stackrel{C}{=}(AB;DX)$ which means $D'\equiv D$ and we are done.

Let $AA_1, BB_1, CC_1$ be the altitudes in $\triangle ABC$ and let $H$ be the orthocenter of $\triangle ABC$. Since $CC_1\perp AB$ and $QP\perp AB$, we get $CC_1\|PQ$. Denote with $\infty$ the point at infinity along $PQ$. Now $$-1 = (Q,P;N,\infty) \overset{C}{=} (A,B;D, C_1).$$ Claim $1$: $D, B_1, A_1$ are collinear. Proof: Let $A_1B_1\cap AB=D_1$. It is enough to show that $D\equiv D_1$. However, from the Ceva/Menelaus picture, we get $$-1=(A, B;D_1, C_1)=(A,B;D, C_1),$$so we are done. Now consider quadrilateral $ABA_1B_1$. Since it is cyclic with center $M$, we can apply Brokard to get that $M$ is the orthocenter of $\triangle CHD$ and thus $H$ is the orthocenter of $\triangle DCM$, so we are done.

Denote $BL \perp AC$, $AK \perp BC$ and $(ABC) \cap (CLK) = Y$. It is well known that M, H, Y lie on one line. Denote $D' = CY \cap AB$. Now H is orthocenter of $\triangle CD'M$ because $CH \perp D'M$ and $MH \perp D'C$. So if we want to prove that the ortocenter of $\triangle CDM$ is H, and we know the ortocenter of $\triangle CD'M$ is H, the only thing we need to prove is that $D \equiv D'$. From N being midpoint of QP we get that (Q, P; N, $P_\infty$) = -1. Projecting through C we get that (Q, P; N, $P_\infty$) = (A, B; D, X) = -1. Now from AYCB cyclic we get that D'Y.D'C = D'A.D'B. From XYCM cyclic (since $\angle MYC = \angle MXC = 90^{\circ}$) we get that D'Y.D'C = D'X.D'M, so D'A.D'B = D'X.D'M. From problem 1 this means that (A, B; D', X) = -1. So we have that (A, B; D, X) = (A, B; D', X) = -1. This means $D \equiv D'$, and that is what we wanted so we are ready.

Let $T$ be the foot of perpendicular from $C$ onto $AB$, $H$ be the orthocenter of $\Delta ABC$ and $P_\infty$ be the point at infinity along $PQ$. $$\implies (P,Q;N,P_\infty) \stackrel{C}{=}(B,A;D,T)=-1$$So $D$ is the $C$ expoint in $\Delta ABC$. Now it suffices to prove $DH \perp CM$ but it is well known that $\overline {D-H-H_c}$ are collinear where $H_c$ is the $C$ humpty point and that $HH_c \perp CM$ and so we are done.

nice projective geometry exercise posting for storage

Let the orthic triangle and orthocenter of $\triangle ABC$ be $\triangle XYZ$ and $H$ respectively. We have that $-1 = (P, Q; N, \infty_{PQ}) \overset{C}= (B, A; D, Z)$ and by Ceva-Menelaus this implies that $D \in XY$. Notice that the altitudes of $C$ onto $AB$ and $C$ onto $MD$ are the same, so it suffices to show that the altitude of $D$ onto $CM$ passes through $H$. However this line is just $DH_C$ where $H_C$ is the $C$-Humpty point and it is well known that $DH_M$ passes through $H$ so we are done.

Let $H$ be the foot from $C$ to $AB$. The desired follows from noting that $D$ is the ex-point, as \[-1 = (PQ;N\infty) \overset{C}{=} (BA;DH). \quad \blacksquare\]