一题:如图,在等边△ABC中,D、F两点分别在AC、BC上,且AF=DC,AD与BF相交与点E.
2个回答

1.

因为等边△ABC 且AF=DC,AB=AC,角BAC=角ACB

所以三角形ABF全等于三角形ADC

所以角DAC=角ABF

又因为角BED=角EBA+角EAB=角DAC+角EAB=角BAC=60度

2.(1)证明:∵DE = BA,∠DBE =∠BCA = 90º,又∠DEB+∠ABC = 90º,∠A+∠ABC = 90º,∴∠DEB =∠A,∴△ACB≌△EBD(AAS),则有BC = BD.

(2)由△ACB≌△EBD,得AC = EB

∵E为BC的中点,

∴EB =1/2BC.

∵BD = 8cm,BC = BD,

∴BC = 8cm.

∴AC = EB =1/2BC = 4cm.

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