[asy]
size(100);
draw((0,0)--(1,0)--(1,1)--(0,1)--cycle);
draw((0,1)--(1,0));
draw((0,0)--(.5,sqrt(3)/2)--(1,0));
label("A",(0,0),SW);
label("B",(1,0),SE);
label("C",(1,1),NE);
label("D",(0,1),NW);
label("E",(.5,sqrt(3)/2),E);
label("F",intersectionpoint((0,0)--(.5,sqrt(3)/2),(0,1)--(1,0)),2W);
//Credit to chezbgone2 for the diagram[/asy]Vertex E of equilateral triangle ABE is in the interior of square ABCD, and F is the point of intersection of diagonal BD and line segment AE. If length AB is 1+3 then the area of △ABF is<spanclass=′latex−bold′>(A)</span>1<spanclass=′latex−bold′>(B)</span>22<spanclass=′latex−bold′>(C)</span>23<spanclass=′latex−bold′>(D)</span>4−23<spanclass=′latex−bold′>(E)</span>21+43