Question

如图,过三角形ABC的边BC上一点D做BC的垂线,交AC和BA的延长线与点E、F,若AE＝AF,试说明AB=AC

Answer 1

AF=AE,

∴∠BFE=∠FEA=∠DEC

∠DEC+∠ECD=90°

∠BFE+∠FBC=90°

又∵∠DEC=∠BFE

∴∠B=∠C

所以AB=AC