(1) verification: eg = CG
(2) Rotate △BEF in Figure ① counterclockwise around point B by 45? As shown in Figure ②, take the G point in DF and connect EG and CG. Is the conclusion in (1) still valid? If yes, please give proof; If not, please explain why.
(3) Rotate the △BEF in Figure ① at any angle around point B, as shown in Figure ③, and then connect the corresponding line segments, and ask whether the conclusion in (1) still holds? What conclusion can you draw from the observation? (No proof required)
Analysis (1) According to the nature of the hypotenuse midline of a right triangle, it is easy to get Ge = GC.
(2) Construct a parallelogram FCDM according to the conditions, connect ME and EC, and easily prove △ MFE △ CBE, thus obtaining GE = GC.
(3) According to the conclusions of (1) and (2), it is inferred that the proof process of GE = GC and EG⊥CG is similar to that of (2).
The solution (1) proves that in Rt△FCD, ∵G is the midpoint of DF,
∴2CG=FD,2EG=FD.
∴CG=EG.
(2) The conclusion in (1) still holds, that is, eg = CG.
Proof: extend CG to m, make mg = CG, and connect MF, ME, EC, DM, FC.
Fg = dg, ∴ quadrilateral MFCD is a parallelogram.
∴MF‖CD‖AB.
In Rt△MFE and Rt△CBE, ∫? MF=CB，EF=BE，
∴△MFE≌△CBE.
∴∠MEC=∠MEF+∠FEC=∠CEB+∠CEF=90。
∴△MEC is a right triangle.
∵? MG=CG，
∴2EG=2CG=MC.
(3) The conclusion in (1) still holds, that is, eg = CG.
Other conclusions are: such as CG.
It shows that the square is the most special quadrilateral, which is a combination of rectangle and diamond. Therefore, when studying quadrilateral, the topic with square as the background is more flexible, representative and comprehensive, which has become a hot topic in previous senior high school entrance examinations. This topic embodies the inherent beauty of harmony in mathematics, the "exploration-guess-proof" process of mathematical problems, and pays attention to the generalization of the conclusion of the topic.