First of all, your painting is wrong. The point G you draw is actually the point F. I set a G point in the second question, which is the vertical foot in your middle.

All four corners of a rectangle are 90 degrees. From ∠ A = 90 and ∠ CEF = 90, we can know that ∠ AEF+∠ CED = 90 and ∠ AEF+∠ AFE = 90, and we can deduce that ∠AFE=∠CED.

The second question is similar to proof. It is proved that the average EF is ∠AFC, ∠AFE=∠CFE, plus ∠A and ∠FEC is 90.

△AFE∽△EFC, combined with the above questions, △AFE∽△EFC∽△DEC, ∠EG⊥CF=∠DCE, and EC shares ∠DCF, which is EG ? CF, then EG=AE(EF shares ∠.