It is proved that the midpoint m of AE is taken to connect AG and GM, and GM is the center line of right-angled trapezoid, so GM⊥AE,

△ AGM △ EGM can be easily proved by SAS, so ∠MGE =∠MGA =∠ Dag =∠DCG.

Let ∠MGE =∠MGA =∠ Dag =∠DCG=∠ 1, then ∠EGB =∠MGB-∠ 1 = 45-∞.

∠CGB=∠BDC+∠GCD=45 +∠ 1，

So ∠ EGC = ∠ EGB+∠ CGB = 45-∠1+45+∠1= 90,

Which is EG vertical CG