如图,在正方形ABCD中,角EAF＝45度,交BC于E,CD于F,AE,AF交正方形对角线于M,N.求证AN垂直于EN.

Question

如图,在正方形ABCD中,角EAF＝45度,交BC于E,CD于F,AE,AF交正方形对角线于M,N.求证AN垂直于EN.

Answer 1

连接AC,
∵四边形ABCD是正方形
∴∠ACE=∠ADN=∠CAD=45°,
∴AC=√2AD,
∵∠EAF=45°,
∴∠EAF=∠CAD,
∴∠EAF-∠CAF=∠CAD-∠CAF,
∴∠EAC=∠NAD,
∴△EAC∽△NAD,
∴AE：AN=AC：AD,
∴AE：AN=√2∶1,
∴AE=√2AN．

余弦定理：EN²=AN²+AE²-2AN×AE×cos45°

                       =AN²+2AN²-2AN×√2AN×√2/2

                       =AN²+2AN²-2AN²

                       =AN²

∴EN=AN

∴∠NEA=∠NAE=45°

∴∠ANE=90°

即AN⊥EN