That is eg = CG. Other conclusions are as follows: ⊥ CG.

Prove:

Extend CG to m so that MG=CG,

Connect MF, ME, ECEF and AB to n

At △DCG? And delta delta △FMG,

FG = DG，∠MGF=∠CGD，MG=CG，

∴△DCG？ ?△FMG。

∴MF=CD,∠FMG=∠DCG.？

∴MF‖CD‖AB.

∴∠MFE=∠ANE=90？ +∞EBA

∫∠EBC = 90？ +∞EBA

∴∠MFE=∠EBC

At △ Maeve and △CEB,

EF=EB？ ∠MFE=∠EBC？ MF=BC

∴△MFE≌△CBE

∴∠MEC=∠MEF+∠FEC=∠BEC+∠FEC=90？

ME=CE

∫G is the midpoint of MC

∴EG=CG (the midline of the hypotenuse of a right triangle is equal to half of the hypotenuse)

EG⊥CG (isosceles triangle connected by three lines)