设底边BC,AB、AC是腰做AD⊥BC于D,AD平分∠BAC,BC∴∠BAD=∠CAD=1/2∠BAC=60°BD=CD=1/2BC=5∴∠ABD=30°AD=1/2AB∴AB²=AD²+BD²=(1/2AB)²+5²3/4AB²=5²AB=10√3/3延长AD交圆于E,连...