Question

数学锐角三角函数点E是矩形ABCD中CD边上一点,△BCE沿BE折叠为△BFE,点F落在AD上.1)求证：△ABF∽△D

Answer 1

（1）由∠AFB+∠ABF=90°,

∠AFB+∠DFE=90°,

∴∠ABF=∠DFE,∠A=∠D=90°

∴△ABF∽△DFE.

（2）由sinDFE=1/3,

设DE=t,EF=3t,DF=2√2t,

∴CE=EF=3t

AF/DE=AB/DF

AF/t=4t/2√2t

∴AF=√2t,

BC=AD=√2+2√2=3√2t,

∴tan∠EBC=EC/BC

=3t/3√2t

=√2/2