Denote these circles as $\omega_{1}$ and $\omega_{2}$ for medians $AA'$ and $BB'$ respectively. Assume that $\omega_{1}$, $\omega_{2}$ interesect at $P$ and $Q$, line $AC$ intersect $\omega _{1}$ at point $K$ and line $BC$ intersect $\omega_{2}$ at point $L$. We have $\angle B'LB = \angle A'KA', $ so triangles $B'LC$ and $KCA'$ are similar. Thus, because points $A'$ and $B'$ are midpoints of sides $BC$ and $AC$ respectively, we have that triangles $LCA$ and $KCB$ are similar. So quadrangle $ABLK$ is cyclic. Thus using radical axis thorem, lines $PQ$, $BL$ and $AK$ are concurrent at point $C$.

Well, little bit faster: We need to show that $C$ has equal powers w.r.t. $\omega_1,\omega_2$; keep the same meaning for $K,L$ from first solution. As shown, $\angle AKA'=\angle BLB'$, consequently $K,L,B',A'$ are concyclic and $LC\cdot CA'=CK\cdot CB'\iff CK\cdot CA=CL\cdot BC$, done. Best regards, sunken rock

Strictly speaking, it should be $\measuredangle AKA'=\measuredangle B'LB$, since $K,L$ may lie on the extension of $CA$ or $CB$.