Solution:

(i) In \[\vartriangle BFD\text{ }and\text{ }\vartriangle BEC,\]

\[\angle BFD\text{ }=\angle BEC\] [Corresponding angles]

\[\angle FBD\text{ }=\angle EBC\] [Common]

Hence, \[\vartriangle BFD\text{ }\sim\text{ }\vartriangle BEC\text{ }by\text{ }AA\]criterion for similarity.

So,

\[BF/BE\text{ }=\text{ }BD/BC\]

\[BF/BE\text{ }=\text{ }{\scriptscriptstyle 1\!/\!{ }_2}\] [Since, D is the mid-point of BC]

\[BE\text{ }=\text{ }2BF\]

\[BF\text{ }=\text{ }FE\text{ }=\text{ }2BF\]

Thus,

\[EF\text{ }=\text{ }FB\]

(ii) In \[\vartriangle AFD,\text{ }EG\text{ }||\text{ }FD\] and using BPT we have

\[AE/EF\text{ }=\text{ }AG/GD\text{ }\ldots .\text{ }\left( 1 \right)\]

Now, \[AE\text{ }=\text{ }EB\][Since, E is the mid-point of AB]

\[AE\text{ }=\text{ }2EF\text{ }\left[ As,\text{ }EF\text{ }=\text{ }FB,\text{ }by\text{ }\left( 1 \right) \right]\]

So, from (1) we have

\[AG/GD\text{ }=\text{ }2/1\]

Therefore, \[AG:\text{ }GD\text{ }=\text{ }2:\text{ }1\]