∴BD=DC

∵⊿∵⊿bdg The circumference of BDG = BD+DG+BG,

Perimeter of quadrilateral ACDG = AC+CD+DG+GA.

∴BG=AG+AC

And BG+ag+AC = ab+AC = b+C.

∴BG=0.5(b+c)

(2) ∵ F is the midpoint of AB and D is the midpoint of BC.

∴BF=0.5AB=0.5c,FD=0.5AC=0.5B，

It is also known from (1) that BF+FG = BG = 0.5 (B+C).

∴FG=0.5b=FD，

∴∠FDG=∠FGD，

It is also easy to draw from the midline theorem:

DE∑AB

∴∠FGD=∠GDE

∴∠FDG=∠GDE

Dg sharing ∠EDF

(3) from ⊿ BDG ∽⊿ DFG ∠ FDG = ∠ B.

It is also known from (2) that < FDG =