四边形ABCD是正方形,G为BC上任意一点（点G与B,C不重合）,AE⊥DG于E,CF∥AE交DG于F.求证：AE=FC+EF.

∵四边形ABCD是正方形,∴AD=DC,∠ADC=90度．又∵AE⊥DG,CF∥AE,∴∠AED=∠DFC=90°,∴∠EAD+∠ADE=∠FDC+∠ADE=90°,∴∠EAD=∠FDC．∴△AED≌△DFC（AAS）．∵△AED≌△DFC,∴AE=DF,ED=FC．∵DF=DE+EF,∴AE=FC+EF．...