As shown in the figure, in RT △ ABC, ∠ a = 90 °, ab = AC = 2, point D is the midpoint of BC side, point E is a moving point on AB side (not coincident with a and b), DF ⊥ de intersects AC at F Let be = x, FC = y （1）DE=DF (2) The functional relation of Y with respect to X and the definition field of X are written out (3) When writing the value of X, EF / / BC

As shown in the figure, in RT △ ABC, ∠ a = 90 °, ab = AC = 2, point D is the midpoint of BC side, point E is a moving point on AB side (not coincident with a and b), DF ⊥ de intersects AC at F Let be = x, FC = y （1）DE=DF (2) The functional relation of Y with respect to X and the definition field of X are written out (3) When writing the value of X, EF / / BC

Through D, do DH ⊥ AB cross AB to h, DN ⊥ AC cross AC to n
So DN ‖ AB, DN = 1 / 2Ab, DH ‖ AC, DH = 1 / 2Ac
So DH = DN, so ∠ ndh = 90 ° because ∠ NDF + ∠ NDE = 90 ° and ∠ NDE + EDH = 90 °
Therefore, EDH = FDN
So △ EDH ≌ △ FDN (ASA)
So de = DF
(2) Because △ EDH ≌ △ fdn
So he = NF
So X-1 / 2Ab = 1 / 2ac-y
That is y = 2-x
Because e is a moving point on the edge of AB (not coincident with a and b)
So x > 0 and X