∵AE is perpendicular to DH, ∴∠ 1+∠ 3 = 90.

∵∠2+∠3=90 ,∴∠ 1=∠2

In right-angle △ADH and right-angle △BAE, ∠ 1=∠2, AD=AB.

∴ Right angle △ADH is equal to right angle △BAE.

DH=AE

2) According to your drawing method, G-point intersection is GN perpendicular to AB, N-point intersection is FM perpendicular to BC, and M-point intersection is F-point intersection.

According to the angle of your mark.

∫GN//BC

∴∠6=∠FEM (internal dislocation angles are equal)

∠∠FEM+∠EFM = 90°，∠6+∠NGH = 90°

∴∠NGH=∠EFM

In right angle △NGH and right angle △MFE, ∠NGH=∠EFM, GN=FM.

∴ Right angle △NGH is equal to right angle △MFE.

∴GH=EF

3) I'll give you a rough idea and try it yourself. Right angle △AHF is similar to right angle △CGE.

AF= 1，DF=3

According to the two vertical lines made in 2), it can be seen that ME= 1, then MF=4, EF= root 17.

That is to say, EF=GH= root number 17.

Then FO=HO= (root number 17)/3.

EO=GO=2 (root number 17)/3

∴ The shaded area is1/2× (17/9)+1/2× (4×17/9) = 85/18.

I do this in my head. Try again yourself. . .