(1) As shown in figure 1, as the center line of BC side, the area of △ABD is equal to the area of △ADC.

(2) As shown in Figure 2, ∠DAB=∠ABC, that is, ABC, point D is on the ray parallel to BC;

(3)① As shown in Figure 3, connecting BD, intersection points A and C are parallel lines A and B of BD respectively, parallel line BF of DC intersects with point F, and the area connecting AF, AC and △AFC is equal to the area of quadrilateral ABCD.

(2) As shown in the figure, connect AC, and after passing through point D, make the extension line of DE∑AC reach point E to BC, and then connect AE.

Because the heights on the common side AC of DE∑AC, △ACD and △AEC are also equal, there is S△ACD=S△AEC.

So s quadrilateral ABCD = s △ ACD+s △ ABC = s △ ABC+s △ AEC = s △ Abe.

So if we take the midpoint f of BE, the straight line AF is the bisector of the area required by the quadrilateral ABCD.