Question

任意三角形ABC,以AB边做等边三角形ABD,再以AC边做等边三角形ACE,连接DC和BE,交点为F,求证:角DFA和角

Answer 1

证明设CD与AB相交于点H

如图,∵∠DAC=60°＋∠BAC,∠BAE=60°＋∠BAC

∴∠DAC=∠BAE

∵AC=AE,AD=AB

∴△ACD≌△AEB

∴∠ADF=∠ACF

∵∠AHD=∠BHF

∴△ADH∽△FBH

∴AH∶HF=DH∶BH

∵∠AHF=∠DHB

∴△AFH∽△DBH

∴∠AFD=∠ABD=60°

同理可得∠AFE=60°

∠AFD=∠AFE