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=AD2、AB=AD也成立,证法与1类似