Sofa and bench were robbed ~! This question is really a bit difficult. Look here specifically. /exercise/math/799580 actually examines the judgment and nature of triangle congruence and the nature of parallel lines, and the practice of auxiliary lines is the key to solving problems.

As shown in the figure, AB=4, rays BM and AB are perpendicular to each other, point D is a moving point on AB, point E is on ray BM, BE= 1/2DB, let EF⊥DE and intercept EF=DE, connect AF, extend intersecting ray BM to point C, let BE=x, BC=y, then the resolution function of y about X.

A.y=- 12x/x-4

B.y=-2x/x- 1

C.y=-3x/x- 1

D.y=-8x/x-4

The process is not complicated or complicated, just think of ideas. Let me tell you the train of thought of this problem. Let FG⊥BC be G, and according to the known conditions, we can get that △DBE is all equal to△ △EGF, FG=BE=x, EG=DB=2x, and then we can get it according to the properties of parallel lines.

I hope to adopt yo ~O(∩_∩)O haha ~