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Junior high school mathematics. Solve. ! ! ! ! .
Solution: (1) af = be. It is proved that in △AFC and △BEC, ∫△ABC and △CEF are equilateral triangles, ∴AC=BC, CF=CE, ∠ ACF = ∠ BCE = 60, which are in △AFC.

AC=BC

∠ACF=∠BCE

CF=CE

∴△AFC≌△BEC(SAS),

∴AF=BE.

2) It holds. Reason: In △AFC and △BEC, ∫△ABC and △CEF are equilateral triangles, ∴AC=BC, CF=CE, ∠ACB=∠FCE=60 degrees, ∴AC b-∞.

AC=BC

∠ACF=∠BCE

CF=CE

∴△AFC≌△BEC(SAS),

∴AF=BE.