仅两题,

仅两题,
仅两题,
 

仅两题,
1、∠BAD=∠EAC=∠EDC,∠AFE=∠DFC
又∠E=180-∠EAC-∠AFE
∠D=180-∠EDC-∠DFC
∴∠E=∠D
∠BAC=∠BAD+∠DAC
∠DAE=∠EAC+∠DAC
∴∠BAC=∠DAE
又AC=AE
∴△AED≌△ACB
∴AB=AD
2、AB=AD也成立,证法与1类似