Solution: In the era of Rt△AFE and△,

Angle FAE= = Angle EAG (AE bisects the bag) Angle AFE= Angle age =90.

AE=AE (public)

So △AFE is equal to △ age (corner edge)

Therefore, EF=EG (the corresponding sides of congruent triangles are equal).

Since DE is perpendicular to BC at point D, D is the midpoint of BC, △BEC is an isosceles triangle, and BE=CE.

In Rt△BFE and △CGE, ef = eg, be = ce.

Then, Rt△BFE is equal to Rt△CGE (HL).

So BF=CG (the corresponding sides of congruent triangles are equal).