Question

正方形ABCD过D点做DE平行AC,角ACE30度CE交AD与点F,求证AE=AF

Answer 1

图自己画

∠DEC=∠ECA=30˚,∠CDE=90˚+∠EDA=90˚+45˚=135˚,

设正方形边长为1,AC=√2,在∆CDE中,用余弦定理,EC/sin135˚=CD/sin30˚, 得EC=√2,所以AC=EC,得

∠CEA=75˚,∠EFA=∠DFC=30˚+45˚=75˚

所以AE=AF