Question

二。三题过程求解

Answer 1

连接KE   KF

∠ACK=∠BCF   

∠ACB=∠BCF+∠ACF=120°   ∠KCF=∠ACK+∠ACF=120°

∠MCN=60°   ∠BCF+∠ACE=60°   ∠KCM=∠ACK+∠ACE=60°

∠KCE=∠ECF=60°  CE=CE  CF=CK

△KCE≌△FCE   得EF=KE

AE^2+EF^2=BF^2   EF=KE  BF=AK

AE^2+KE^2=AK^2    有 KE⊥AE

而∠KAE=30°+30°   AK=2/根号3*KE  

KE⊥AE  EF=KE  根号2/2*KF=KE  

∠KCF=∠ACK+∠ACF=120°   CF=CK

KF=根号3*CF

BF=AK=2/根号3*KE =2/根号3*根号2/2*根号3*CF

即  BF=根号2*CF

AB=根号3*AC

AE=根号3/3*EF    BF=AK=2/根号3*EF

根号3/3*E+2/根号3*EF+EF=根号3*（根号3+1）

EF=（根号3+1）/（3+根号3）=根号3/3