证明：过B作角平分线BD,与AC交于D,则∵∠ABC=2∠ACB,BC=2AB,∴∠DBC=∠ABC,∴△DBC是等腰三角形,过D作DE⊥BC于E,则E同时为BC的中点,即EB=EC=(1/2)BC=AB,∵DB=DB,∠DBC=∠DBA,∴△DBE≌△DBA,∴∠DAB=∠DEB=90°即∠B...