Solution:
Let $P(A, B, C)$ be the problem assertion. First consider a non-right triangle $A B C$ with orthocenter $H$. Note that in the set $\{A, B, C, H\}$, the last point is the orthocenter of the other three. Thus, we can assume WLOG $f(A) \leq f(B) \leq f(C) \leq f(H)$.
By considering $P(A, B, C)$, this implies $f(A)+f(C)=f(B)+f(H)$. Since $f(A) \leq f(B)$ and $f(C) \leq f(H)$, this implies equality must hold in both inequalities, and so $f(A)=f(B)$ and $f(C)=f(H)$.
Denote by $\Omega_{B C}$ the circle with diameter $B C$. We prove the following claim.

Claim: If $f(B) \neq f(C)$, then for all $D \in \Omega_{B C}$, we have $2 f(D)=f(B)+f(C)$. In particular, $f$ is constant on $\Omega_{B C}$.

Proof: We split into cases. If $f(D) \notin [f(B), f(C)]$, then $P(B, C, D)$ implies
$$f(D)+f(C)=f(B)+f(D)$$
which implies $f(B)=f(C)$, contradiction. Thus, $f(D) \in [f(B), f(C)]$, from which $P(B, C, D)$ implies
$$f(B)+f(C)=2 f(D)$$
The claim then follows.

Now we claim that $f(B)=f(C)$. Assume for the sake of contradiction that this was not the case. Consider $\Omega_{A C}$, and let $D=A H \cap B C \in \Omega_{A C}$. Let $\Omega_{B D'}$ intersect $\Omega_{A C}$ at a second point $D'' \neq D'$. Then since $f(A) \neq f(C)$, from the claim we get $2 f\left(D'\right)=f(A)+f(C)$.

Now we have two cases. In the first case, suppose $f(B)=f\left(D'\right)$. Then $2 f(B)=f(A)+f(C)$, implying $f(B)=f(C)$, contradiction. In the second case, we have $f(B)+f\left(D'\right)=2 f\left(D''\right)=f(C)+f(A)$. This implies $f(C)=f\left(D'\right)$, and so
$$2 f(C)=2 f\left(D'\right)=f(C)+f(A)$$
implying $f(C)=f(A)=f(B)$, another contradiction. Thus our assumption was wrong, and $f(B)=f(C)$. This implies $f(A)=f(B)=f(C)=f(H)$ for any nondegenerate non-right triangle $A B C$, and so $f$ is constant everywhere (by considering two segments whose diameter circles do not intersect). It is clear that these solutions work.