We first construct the point $O$ of the intersection of diagonals, and the point $T$ of intersection of sides $AB$ and $CD$. It is well known that $OT$ contains midpoints $P$ and $E$ of parallel sides (fig. 11). We have $AE = ED = BC = a$. Hence $ABCE$ and $BCDE$ are parallelograms, $M$ and $N$ are the points of intersection of their diagonals respectively. We can draw the line $MN$ which is a midline of trapezoid and intersects two sides at $K$ and $L$ respectively. We then draw line $BL$, that intersects $AD$ at point $F$. Thus $\square ABF$ is desired, since $\square ABF = \square BCL$.