如图,在RT△ABC中和RT△ADE中,∠C=∠E=90°,BC=DE,∠BAE=∠DAC,BC与DE交于点F,求证：BF=DF

连接AF∵∠BAE=∠DAC∴∠BAE+∠EAC=∠DAC+∠EAC即∠BAC=∠DAE在△AED与△ACB中∠C=∠E∠BAC=∠DAEBC=DE∴△AEF≡△ABC(A.A.S)∴AE=AC CB=ED在△AEF与△ACF中AF=AFAE=AC∴△AEF≌△ACF（H.L）∴FE=CF∵BF=CB-CFDF=DE-...