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

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．