Question

如图,E、F是菱形ABCD边BC、CD上的点,且∠EAF=∠D=60 ,求证BE=BF

Answer 1

(1)连接AC,易知∠CAB=1/2 BAD=60° =∠ACF

△ABC为等边三角形

∵∠CAF+∠CAE=∠EAF=60°

∠BAE+∠CAE=∠CAB=60°

∴∠BAE=∠CAF

又 ∵∠EBA=∠FCA=60°

且AC=AB

∴△EBA≌△FCA

所以AE=AF

(2)∵AE=AF,∠EAF=60°

所以△EAF为等边△

∠AFE=60°

∠AFD=180°-∠FAD-∠D=180°-60°-45°=75°

所以∠CFE=180°-∠AFD-∠AFE=180°-75°-60°=45°