Question

如图,已知Rt△SBC≌Rt△ADE,∠ABC=∠ADE=90°,BC与DE相交于点F连接CD,EB.求证CF=EF

Answer 1

证法一：连接CE,

∵Rt△ABC≌Rt△ADE,

∴AC=AE．

∴∠ACE=∠AEC．

又∵Rt△ABC≌Rt△ADE,

∴∠ACB=∠AED．

∴∠ACE=∠ACB=∠AEC-∠AED．

即∠BCE=∠DEC．

∴CF=EF．

证法二：∵Rt△ABC≌Rt△ADE,

∴AC=AE,AD=AB,∠CAB=∠EAD,

∴∠CAB-∠DAB=∠EAD-∠DAB．

即∠CAD=∠EAB．

∴CD=EB,∠ADC=∠ABE．

又∵∠ADE=∠ABC,

∴∠CDF=∠EBF．

又∵∠DFC=∠BFE,

∴△CDF≌△EBF．

∴CF=EF．

证法三：连接AF,

∵Rt△ABC≌Rt△ADE,

∴AB=AD,BC=DE,∠ABC=∠ADE=90°．

又∵AF=AF,

∴Rt△ABF≌Rt△ADF（HL）．

∴BF=DF．

又∵BC=DE,

∴BC-BF=DE-DF．

即CF=EF．