在正方形ABCD中,E为CD的中点,F为BC上一点,且CF=(1/4)BC,
试说明AE⊥EF.
因为,在△ADE和△ECF中,
∠ADE = 90°= ∠ECF ,AD/DE = 2 = EC/CF ,
所以,△ADE ∽ △ECF ,
可得：∠DAE = ∠CEF ,
所以,∠AEF = 180°-∠AED-∠CEF = 180°-∠AED-∠DAE = ∠ADE = 90°,
即有：AE⊥EF .