一道几何解答题
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设CP于圆相交于点M,连接FM,作MH⊥CD于点H
则由∠FPM=90°,可知FM为直径,∴FM⊥AF
即FM//BC,∴∠MCH=∠PMF。设圆O半径=r
则BD=BF=MH=r,FM=2r,设tan∠PMF=x>0,则
CH=MH/tan∠MCH=r/x,∴CD=CE=DH+HC=r(x+1)/x
连接FD,可知∠FDB=45°,∴tan∠ADB=tan(45°+∠FDP)
=tan(45°+∠FMP)=(1+tan∠FMP)/(1-tan∠FMP)=(1+x)/(1-x)
∴AB=BDtan∠ADB=r(1+x)/(1-x) => AF=AE=AB-BF=2xr/(1-x)
再由BC²+AB²=AC²,而BC=BD+CD,AB=AF+FB,AC=AE+EC
∴(BD+DC)²+(AF+FB)²=(AE+EC)²,注意AF=AE,CD=CE,BD=BF
=>BD²+2BD*DC+DC²+AF²+2AF*FB+FB²=AE²+2AE*EC+EC²
=>BD²+BD*(AE+EC)=AE*EC,带入BD,AE,EC的值即得
r²+r[2xr/(1-x)+r(x+1)/x]=2r²(1+x)/(1-x) => 2x²+x-1=0
∵x>0,∴解得x=1/2 =>CD=r(x+1)/x=3r,AF=2xr/(1-x)=2r
即AB=3r,BC=4r,AC=5r
∴三边之比AB:BC:AC=3:4:5
则由∠FPM=90°,可知FM为直径,∴FM⊥AF
即FM//BC,∴∠MCH=∠PMF。设圆O半径=r
则BD=BF=MH=r,FM=2r,设tan∠PMF=x>0,则
CH=MH/tan∠MCH=r/x,∴CD=CE=DH+HC=r(x+1)/x
连接FD,可知∠FDB=45°,∴tan∠ADB=tan(45°+∠FDP)
=tan(45°+∠FMP)=(1+tan∠FMP)/(1-tan∠FMP)=(1+x)/(1-x)
∴AB=BDtan∠ADB=r(1+x)/(1-x) => AF=AE=AB-BF=2xr/(1-x)
再由BC²+AB²=AC²,而BC=BD+CD,AB=AF+FB,AC=AE+EC
∴(BD+DC)²+(AF+FB)²=(AE+EC)²,注意AF=AE,CD=CE,BD=BF
=>BD²+2BD*DC+DC²+AF²+2AF*FB+FB²=AE²+2AE*EC+EC²
=>BD²+BD*(AE+EC)=AE*EC,带入BD,AE,EC的值即得
r²+r[2xr/(1-x)+r(x+1)/x]=2r²(1+x)/(1-x) => 2x²+x-1=0
∵x>0,∴解得x=1/2 =>CD=r(x+1)/x=3r,AF=2xr/(1-x)=2r
即AB=3r,BC=4r,AC=5r
∴三边之比AB:BC:AC=3:4:5
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