∫△EGH?△BCF，

∴∠DCB=∠G=90，FC=GH，

∴FC∥GH，

∴ Quadrilateral FCGH is a parallelogram,

The quadrangle FCGH is a rectangle,

∴ Two moving points E and F start from vertices B and C respectively and move at the same speed on edges BC and CD.

∴BE=CF= 1

∫ In the rectangular ABCD, the side length AB=3, and Tan ∠ Abd = 43.

∴BC=4

∴EC=3

EG = BC

∴CG= 1

∴CG=CF，

∴ Quadrilateral CGHF is a square.

∴DF=2？ FH= 1

∴dh=5；

(2) Make the areas of △BOE and △DOF equal. As can be seen from the figure, as long as the areas of △BCF and △DCE are equal.

∫SBCF = 12×BC×CF，SDCE= 12×CE×CD，

∵ According to (1), CF=BE, △EGH always keeps △ egh △ BCF during the exercise.

∴CF=BE=4-CE will not change,

∴BC×BE=(4-BE)×CD

∴ Substitution value be =127;

(3) From the meaning of the question, DM=DE

Perpendicular bisector whose CD is EM.

Knowing FH∑BC from (1)

∴FHCM=DFDC

∫FH = BE = FC CE = BC-BE

∴BEBC？ BE=CD？ BECD

Is the substitution value BE4? BE=3？ BE3，

Is the solution be = 5? 13.