In a square ABCD, AB=2, E is a point on the line segment CD (point E does not coincide with points C and D), FG bisects AE vertically, AE intersects with F, and AB intersects with an extension line at

In a square ABCD, AB=2, E is a point on the line segment CD (point E does not coincide with points C and D), FG bisects AE vertically, AE intersects with F, and AB intersects with an extension line at G. Solution: 1, if there is a problem, you can get ∠ d = ∠ AFG = 90, ∠eag=∠dea, so: △ADE~△GFA.

2, the similarity of two triangles from 1, ag/ae=af/de. Ag=2+y, ae=[ root sign (x? +2? )],

Af=[ root symbol (x? +2? ) ]/2, de = X. You can get y=[(x-2)? ]/(2x). Domain (2 > x > 0) As for the scope of this problem, it is not necessary to consider that it is greater than 0, and other problems should consider that Y is greater than 0.

3. It can be proved that △ade~△gbh. Get bh/de=gb/ad. And the next day's y=[(x-2)? ]/(2x).X= 1 solvable, that is, de= 1.

Note: the process of solving the answer in the middle is not written, and you can get the result by substituting the data yourself.