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).