试用仿射坐标系来解决
以B为原点，BC和BA为两条轴及单位长度建立仿射坐标系，则三点坐标为
A(0,1), B(0,0), C(1,0)
再设D点坐标为 D(a,1), 则E点坐标为 E((a+1)/2, 1/2)
现来求O点作为AE和BD交点的坐标
直线AE: x/((a+1)/2) = (y-1)/(1/2-1)
BD: x = ay
解得 x = a(a+1)/(2a+1)
y = (a+1)/(2a+1)
因此O点坐标是
O(a(a+1)/(2a+1), (a+1)/(2a+1))
现在可用以行列式
x1, y1, 1
x2, y2, 1
x3, y3, 1
求几个三角形在仿射坐标系下的面积：
△ABO：
0, 1, 1
0 , 0, 1
a(a+1)/(2a+1), (a+1)/(2a+1), 1
ABO = a(a+1)/(2a+1)
△OED:
a(a+1)/(2a+1), (a+1)/(2a+1), 1
(a+1)/2, 1/2, 1
a, 1, 1
OED = a(-a)/(2a+1), -a/(2a+1), 0
1/2-a/2, -1/2, 0
a, 1, 1
= 1/2*a^2/(2a+1) + a/(2a+1)(1/2-a/2)
= a/(4a+2)
△ABC：
0, 1, 1
0, 0, 1
1, 0, 1
ABC = 1
△ACD:
0, 1, 1
1, 0, 1
a, 1, 1
ACD = a
注意仿射变换下面积值会变， 因此以上得出的ABO面积a(a+1)/(2a+1)并不等同于题设的40. 好在两个面积的比值在仿射变换下不变，于是有
40/12 = ABO/OED = a(a+1)/(2a+1) * (4a+2)/a = 2(a+1)
得 a = 2/3
ABCD/ABO = (ABC+ACD)/ABO= (a+1)*(2a+1)/(a(a+1)) = (2a+1)/a = (2*2/3+1)/(2/3) = 7/2
所求梯形面积为 7/2 * S_ABO = 7/2 * 40 = 140
