如图,AB为圆O的直径,PC切圆o于C交BA延长线于p,BD⊥PC于B,
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如图,AB为圆O的直径,PC切圆o于C交BA延长线于p,BD⊥PC于B,
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=1:3
OP=3r;
OC:BD=OP:BP
r: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:BA
BC²=(4/3)r*2r=(8/3)r²,BC=r(2/3)√6
AC²=AB²-BD²=4r²-(8/3)r²=(4/3)r²,AC=r(2/3)√3;
AC+BC=2+2√2
r(2/3)√3+r(2/3)√6=2+2√2
r=(3+3√2)/(√3+√6)=(3+3√2)(√6-√3)/[(√3+√6)(√6-√3)]=(3√6-3√3+3√12-3√6)/(6-3)=√3;
CD²=BC²-BD²=(8/3)r²-[(4/3)r]²=(8/9)r²
CD=r(2/3)√2;
PD=CD+CP=4CD=r(8/3)√2;
SRT△PBD=BD*PD/2=(4/3)r*r(8/3)√2/2=(16/9)*3*√2=(16/3)√2.
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=1:3
OP=3r;
OC:BD=OP:BP
r: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:BA
BC²=(4/3)r*2r=(8/3)r²,BC=r(2/3)√6
AC²=AB²-BD²=4r²-(8/3)r²=(4/3)r²,AC=r(2/3)√3;
AC+BC=2+2√2
r(2/3)√3+r(2/3)√6=2+2√2
r=(3+3√2)/(√3+√6)=(3+3√2)(√6-√3)/[(√3+√6)(√6-√3)]=(3√6-3√3+3√12-3√6)/(6-3)=√3;
CD²=BC²-BD²=(8/3)r²-[(4/3)r]²=(8/9)r²
CD=r(2/3)√2;
PD=CD+CP=4CD=r(8/3)√2;
SRT△PBD=BD*PD/2=(4/3)r*r(8/3)√2/2=(16/9)*3*√2=(16/3)√2.
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