已知Rt三角形ABC全等于Rt三角形ADE,角ABC=角ADE=90度,BC与DE相交与点F,连接CD,EB

△ADC≌△ABE,△CDF≌△EBF；证明：∵Rt△ABC≌Rt△ADE,∴AC=AE,AD=AB,∠CAB=∠EAD,∴∠CAB-∠DAB=∠EAD-∠DAB．即∠CAD=∠EAB．∴△ADC≌△ABE,∴CD=EB,∠ADC=∠ABE．又∵∠ADE=∠ABC,∴∠CDF=∠EBF．又∵∠DFC=∠B...第二对是边角边吗哪来的叫ADE等于叫ABC∠ADE、∠ABC两个角全是直角呀，当然相等