(1) Because △ABC and △ACD have the same base (CD=BC) and the same height (which is the vertical line segment from point A to BD), the area of △ACD is equal to that of △ABC, so s1= a. 。

(2) Because of the expansion and extension of this problem, if AD is connected, we can get S△ABC=S△ACD=S△AED from the conclusion of the above problem.

Therefore, S2=S△ABC+ S△AED=a+a=2a.

(3) From the conclusions of (1) and (2), it can be concluded that S△AEF= S△ACD= S△BDF=2a.

S3=6a。

Find:

It is easy to get from (1), (2) and (3), and the area of △DEF extended outward once is 7 times that of △ABC. The area of the triangle expanded twice is 49 times that of △ABC, and the area of the triangle expanded n times is 7n times that of △ABC.

Application:

From the previous problem solving process, it is found that the area of △GHM is 72 times that of △ABC, and the area of shadow part is (72- 1)* 10=480(m2).