Approach to part ii:

1. Angle ADB = 90 degrees (from part i).

2. AB = AC (given).

3. Triangle ABC is isoceles (by 2).

4. BD = DC (by 1 & 3).

5. EG = GB (given).

6. DG // CE (by mid-point theorem).

Approach to part iii:

1. CE // DG (from part ii).

2. FE // DG (by 1).

3. AE = EG (given).

4. AF = FD (by intercept theorem).

So AF = 1/2 AD.