Solution. Naturally, $g$ and $h$ are the $A$ and $H$-symmedians of $\bigtriangleup ABC$ and $\bigtriangleup AHB$, respectively. Since $A'B'$ is antiparallel to $AB$, it follows that both $g$ and $h$ pass through the midpoint of $A'B'$. $\square$

Here is a projective solution if you, like me, don't know what anti-parallel means. Let $\Gamma$ be the circumcircle of $\triangle ABC$. We use directed angles modulo $180^\circ$. For convenience, let $A'=E$ and $B'=F$. Claim 1. $h$ bisects $EF$ in $M$. Proof. Since $AFEB$ is cyclic, we have $\angle FEH=\angle FEA=\angle FBA$. As $\angle AHB=\angle EHF$, we see that $\triangle AHB$ and $\triangle EHF$ are oppositely similar. As $h$ is a symmedian in $\triangle AHB$, it is a median in $\triangle EHF$. $\blacksquare$ Now it is enough to show that $g$ passes through $M$. However, $g$ is the $C$-symmdian, so $X=g\cap \Gamma$ induces a harmonic quadrilateral $AXBC$. Now define $\ell$ to be the tangent to Gamma at $C$. Claim 2. $\ell\parallel EF$ Proof. \[\angle (\ell,BC)=\angle BAC=\angle BAF=\angle BEF=\angle CEF.\quad \blacksquare\] Projecting $AXBC$ onto line $EF$ from $C$, we get \[-1=(A,B;X,C)=(F,E;g\cap EF,P_{\infty}).\]Thus, $g\cap EF=M$, as desired.

$\color{blue}\boxed{\textbf{SOLUTION}}$ $g$ is the symmedian of $\triangle ABC$ and $A'B'$ is antiparallel to $AB$ So, $g$ pass through the midpoint of $A'B'$ call $K$ We need to show $h$ pass through the midpoint of $A'B',K$ that is $\angle AHC_0=\angle BHK$ We have, $\triangle ABH$ is similar to $\triangle A'B'H$ $C_0,K$ be the midpoints of $AB$ and $A'B'$ respectively $\implies \triangle AHC_0$ is similar to $\triangle BHK \implies \angle AHC_0=\angle BHK \blacksquare$

Is there no solution that does not require knowledge of the symmedian?