解：∵$AD$为$∆ABC$的高，∴$∠C + ∠CAD = 90°$
$ $又$∠C = 40°，$∴$∠CAD = 50°$
$ $又$∠DAE = 12°，$$∠CAD = ∠DAE + ∠CAE$
∴$∠CAE = 38°$
$ $又$AE$平分$∠BAC，$$BF $平分$∠ABC$
∴$∠BAC = 2∠CAE = 76°，$$∠CBF = \frac 12∠ABC$
$ $又$∠ABC + ∠BAC + ∠C = 180°$
∴$∠ABC = 180° - ∠BAC - ∠C = 64°，$即$∠CBF = 32°$
$ $又$∠BF C + ∠C + ∠CBF = 180°$
∴$∠BF C = 180° - ∠C - ∠CBF = 108°$
$ $又$∆GF C$为直角三角形
∴$∠FG C = 90°$或$∠GF C = 90°。$分类讨论如下：
$ ① $当$∠FG C = 90°$时
∵$∠FG C = ∠CBF + ∠BFG$
∴$∠BFG = 90° - ∠CBF = 58°$
$ ② $当$∠GF C = 90°$时，
$∠BFG = ∠BF C - ∠GF C = 18°$
综上，$∠BFG $的度数为$58°$或$18°$
