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BD⊥PC于D?PC切圆O于C,连接OC,则OC⊥PC于C,设圆O的半径为r，OC//BD,OB:OP=CD:CP=1：3;CP=3CD;r:OP=1:3OP=3r;OC:BD=OP:BPr:BD=3r：(BO+OP)BD=r(r+3r)/(3r)=(4/3)r;∠BCD=∠BAC，【弦切角=同弧上的圆周角】RT△BDC∽RT△BCA,[AA]BD:BC=BC:BABC2=(4/3)r*2r=(8/3)r2,BC=r(2/3)√6AC2=AB2-BD2=4r2-(8/3)r2=(4/3)r2,AC=r(2/3)√3;A... |
