已知等腰△ABC,PB=3,PC的平方=7,AP=1,∠A=90°,则AB=AC,设AB=AC=M过P点作PD⊥AC,交AC于D点,作PE⊥AB,交AB于E点,则四边形PDAE为矩形,PE=DA=X,PD=EA=Y,CD=AC-DA=M-X根据勾股定理,得PA^2=DA^2+PD^2,1=X^2+Y^2.(1)PC^2=P...