已知,在△ABC中,∠CAB=2∠B,AE平分∠CAB交BC于E,AB=2AC,求证：1）∠C=90°

证明：∵AE平分∠CAB,∴∠CAB=2∠BAE,
∵∠CAB=2∠B,∴∠EAB=∠B,∴EA=EB,
过E作ED⊥AB于D,则AB=2AD,
∵AB=2AC,∴AD=AC,
在ΔAED与ΔAEC中,
AD=AC,∠EAD=∠EAC,AE=AE,
∴ΔAED≌ΔADC,
∴∠C=∠ADE=90°.