需要求证的结论应该是:CF·CA=CF·EA.
[证明]
∵BD是直径,∴AE⊥DE,∴∠AEC=90°-∠DEF.
∵B、D、F、E共圆,∴∠DEF=∠DBG,∴∠AEC=90°-∠DBG.
∵BC⊥CG,∴∠FGC=90°-∠DBG,∴∠AEC=∠FGC.······①
∵AE⊥DE、AC⊥DC,∴A、E、D、C共圆,∴∠EAC=∠BDE.
∵B、D、F、E共圆,∴∠BDE=∠BFE,∴∠EAC=∠BFE,而∠BFE=∠GFC,
∴∠EAC=∠GFC.······②
由①、②,得:△AEC∽△FGC,∴EA/GF=CA/CF,∴CF·CA=CF·EA.
[以下证明CF>CD.]
∵B、D、F、E共圆,∴∠CDF=∠BEF>∠BED=90°.
∵B、D、F、E共圆,∴∠CFD=∠ABC<90°,∴∠CDF>∠CFD,∴CF>CD.