如图平行四边形ABCD中AE,AF是高∠BAE=30°,BE=2 CF=1,DE交AF于点G...答:(1)解:AE垂直BC,∠BAE=30°,则AB=2BE=4=DC,AE=2√3;DF=DC-CF=3.AF垂直CD,∠ADF=∠B=60°,则∠DAF=30°,AD=2DF=6=BC,CE=4.S△ECD=CE*AE/2=4√3;(2)证明:DE=√(AD^2+AE^2)=√(36+12)=4√3.则AE=DE/2;又AE垂直BC,BC平行AD,则AE垂直AD,得∠ADE=30°.∠AGE=...