[asy][asy] size(300); unitsize(0.75); defaultpen(linewidth(0.8)); pair A=origin, B=(200,0), C=(133,77); draw(A--B--C--cycle); path LLeg=C--A; path RLeg=C--B; pair D=waypoint(LLeg,1/3), E=waypoint(LLeg,2/3), F=waypoint(RLeg,1/3), G=waypoint(RLeg,2/3); dot(D); dot(E); dot(F); dot(G); label("$A$",A,W); label("$B$",B,dir(0)); label("$C$",C,N); label("$D$",D,NW); label("$E$",E,NW); label("$F$",F,NE); label("$G$",G,NE); draw(E--F^^D--G); pair [] H=intersectionpoints(E--F,D--G); dot(H[0]); label("$H$",H[0],dir(280)); pair M=waypoint(A--B,1/2); dot(M); label("$M$",M,S); draw(M--C); [/asy][/asy] Suppose we want to bisect $\overline{AB}.$ Then draw in an arbitrary point $C$ and trisect $\overline{AC}$ and $\overline{BC}$ using points $D,E,F,$ and $G.$ Draw $\overline{DG}$ and $\overline{EF}$ and label their intersection $H.$ The intersection of line $\overleftrightarrow{CH}$ with $\overline{AB}$ is $M,$ the midpoint of $\overline{AB}.$ We are done. To see why this works, consider $\triangle EGC.$ Its medians $\overline{DG}$ and $\overline{EF}$ intersect at $H,$ so $H$ must be the centroid of $\triangle EGC.$ This implies that the extension of $\overline{CH}$ is a median of $\triangle EGC$ since it passes through the centroid, so by definition it bisects $\overline{EG}.$ Since $\triangle EGC\sim\triangle ABC,$ we can conclude that it bisects $\overline{AB}$ as well.