Another solution with four triangles later incorporated into the question. Start with square $ABCD$. From$A$ draw a line segment to a point $E$ on side $BC$. Construct the perpendicular to $AE$ through $E$ which intersects side $CD$ at $F$. Finish the division by drawing $AF$. This must be distinct from the cases given above because only one vertex of the square is used for triangle vertices with acute angles.