It is known that the four vertices of a square are o (0,0), a (1,0), B (1,1), C (0,1), and points D and e move on line OC and ab respectively, and OD = be. Let AD and OE intersect at point G, then the trajectory equation of point G is () A. y=x（1-x）（0≤x≤1）B. x=y（1-y）（0≤y≤1）C. y=x2（0≤x≤1）D. y=1-x2（0≤x≤1）

It is known that the four vertices of a square are o (0,0), a (1,0), B (1,1), C (0,1), and points D and e move on line OC and ab respectively, and OD = be. Let AD and OE intersect at point G, then the trajectory equation of point G is () A. y=x（1-x）（0≤x≤1）B. x=y（1-y）（0≤y≤1）C. y=x2（0≤x≤1）D. y=1-x2（0≤x≤1）

Let D (0, m) (0 ≤ m ≤ 1), then E (1, 1-m), so the equation of straight line ad is x + YM = 1, the equation of straight line OE is y = (1-m) x, let g (x, y), then by X + YM = 1y = (1 − m) x, we can get x = my = (1 − m) m, by eliminating m, we can get y = (1-x) x (0 ≤ x ≤ 1)