Question

已知AD‖BC,BD=CD,∠DBC=45°,BD⊥DC,CE⊥AB,求证:CF=AB+AF

Answer 1

证明：在线段CF上截取CH=BA,连接DH,

∵BD⊥CD,BE⊥CE,

∴∠EBF+∠EFB=90°,∠DFC+∠DCF=90°,

∵∠EFB=∠DFC,

∴∠EBF=∠DCF,

∵DB=CD,BA=CH,

∴△ABD≌△HCD,

∴AD=DH,∠ADB=∠HDC,

∵AD∥BC,

∴∠ADB=∠DBC=45°,

∴∠HDC=45°,∴∠HDB=∠BDC-∠HDC=45°,

∴∠ADB=∠HDB,

∵AD=HD,DF=DF,

∴△ADF≌△HDF,

∴AF=HF,

∴CF=CH+HF=AB+AF,

∴CF=AB+AF．