Point F is taken on the extension of side AD of parallelogram ABCD. BF intersects diagonal AC at E and side DC at G. If EF=32 and GF=24, then BE equals:
[asy]
size(7cm);
pair A = (0, 0), B = (7, 0), C = (10, 5), D = (3, 5), F = (5.7, 9.5);
pair G = intersectionpoints(B--F, D--C)[0];
pair E = intersectionpoints(A--C, B--F)[0];
draw(A--D--C--B--cycle);
draw(A--C);
draw(D--F--B);
label("A", A, SW);
label("B", B, SE);
label("C", C, NE);
label("D", D, NW);
label("F", F, N);
label("G", G, NE);
label("E", E, SE);
//Credit to MSTang for the asymptote
[/asy]<spanclass=′latex−bold′>(A)</span>4<spanclass=′latex−bold′>(B)</span>8<spanclass=′latex−bold′>(C)</span>10<spanclass=′latex−bold′>(D)</span>12<spanclass=′latex−bold′>(E)</span>16