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電子課本網(wǎng) 第75頁(yè)

第75頁(yè)

信息發(fā)布者:
證明:?$(1)∵C$?是?$\widehat{BD}$?的中點(diǎn),
?$∴\widehat{CD}=\widehat{BC}.$?
?$∵AB$?是?$⊙O$?的直徑,且?$CF⊥AB,$?
?$∴\widehat{BC}=\widehat{BF},$?
?$∴\widehat{CD}=\widehat{BF},$?
?$∴CD=BF.$?
∵在?$△BFG $?和?$△CDG $?中,?$∠F=∠CDG,$??$∠FGB=∠DGC,$??$BF=CD,$?
?$∴△BFG≌△CDG.$?
?$(2)$?如圖,過(guò)?$C$?作?$CH⊥AD$?于?$H,$?連接?$AC、$??$BC,$?
?$∵\(yùn)widehat{CD}=\widehat{BC},$?
?$∴∠HAC=∠BAC.$?
?$∵CE⊥AB,$?
?$∴CH=CE.$?
?$∵AC=AC,$?
?$∴Rt△AHC≌Rt△AEC,$?
?$∴AE=AH.$?
?$∵CH=CE,$??$CD=CB,$?
?$∴Rt△CDH≌Rt△CBE,$?
?$∴DH=BE=2,$?
?$∴AE=AH=2+2=4,$?
?$∴AB=4+2=6.$?
?$∵AB$?是?$⊙O$?的直徑,
?$∴∠ACB=90°,$?
?$∴∠ACB=∠BEC=90°.$?
?$∵∠EBC=∠ABC,$?
?$∴△BEC∽△BCA,$?
?$∴\frac {BC}{AB}=\frac {BE}{BC},$?
?$∴BC^2=AB·BE=6×2=12,$?
?$∴BF=BC=2\sqrt{3}.$?

B
C
135°
證明?$:(1)$?連接?$OD. $?
?$∵ AB$?是?$⊙O$?的直徑, 
?$∴ ∠ADB=90°,$?
∴ 在?$Rt△ADB$?中?$,∠ABD+∠OAD=90° $?
?$∵ OA=OD,$?
?$∴ ∠OAD=∠ODA. $?
?$∵ ∠CDF=∠ABD, $?
?$∴ ∠ODA+∠CDF=90°, $?
?$∴ ∠ODF=90°,$? 
?$∴ DF⊥OD.$?
?$∵ OD$?是?$⊙O$?的半徑, 
?$∴ DF $?是?$⊙O$?的切線
?$(2)$?連接?$AE. $?
?$∵ BE=DE,$?
?$∴ ∠BAE=∠CAE. $?
?$∵ AB$?是?$⊙O$?的直徑, 
?$∴ ∠AEB=90°=∠AEC, $?
?$∴ ∠ABE+∠BAE=90°,∠ACE+∠CAE=90°,$?
?$∴ ∠ABE=∠ACE, $?
?$∴ AB=AC.$?
根據(jù)?$ \frac {AD}{BD} = \frac {4}{3} ,$?
設(shè)?$AD=4x,$?
則?$BD=3x,$? 
∴ 在?$Rt△ABD$?中,由勾股定理,得?$AB=\sqrt{(4x)2+(3x)2} =5x,$? 
?$∴ AC=5x,$? 
?$∴ CD=x. $?
∵ 在?$Rt△BDC$?中?$,BD2+CD2=BC2, $?
?$∴ (3x)2+x2=( \sqrt{10} )2,$?
解得?$x=1($?負(fù)值舍去),
?$ ∴ AB=5x=5,$? 
?$∴ ⊙O$?的半徑為?$ \frac {1}{2}AB= \frac {5}{2}$

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