人教版九年級(jí)數(shù)學(xué)上冊(cè) 專項(xiàng)素養(yǎng)綜合全練(五)（含答案解析）

專項(xiàng)素養(yǎng)綜合全練(五)圓的切線的證明類型一見(jiàn)半徑,證垂直1.如圖,在△ABC中,AB=AC,以AB為直徑作☉O,過(guò)點(diǎn)O作OD∥BC交AC于D,∠ODA=45°.求證:AC是☉O的切線.如圖,AB是☉O的直徑,點(diǎn)C為☉O外一點(diǎn),連接OC交☉O于點(diǎn)D,連接BD并延長(zhǎng)交線段AC于點(diǎn)E,∠CDE=∠CAD,求證:AC與☉O相切.類型二連半徑,證垂直3.如圖,AB為☉O的直徑,AC平分∠BAD交☉O于點(diǎn)C,CD⊥AD,垂足為點(diǎn)D.求證:CD是☉O的切線.4.如圖,以點(diǎn)O為圓心,AB長(zhǎng)為直徑作圓,在☉O上取一點(diǎn)C,延長(zhǎng)AB至點(diǎn)D,連接DC,∠DCB=∠DAC,過(guò)點(diǎn)A作AE⊥AD交DC的延長(zhǎng)線于點(diǎn)E.(1)求證:CD是☉O的切線;(2)若CD=4,DB=2,求AE的長(zhǎng).5.如圖,在△ABC中,AB=AC,☉O是△ABC的外接圓,作∠BCD=∠ACB,連接AD交BC于點(diǎn)E,延長(zhǎng)DC至點(diǎn)F,使CF=AC,連接AF.(1)求證:ED=EC;(2)求證:AF是☉O的切線.6.如圖,在△ABC中,D為BC邊上的一點(diǎn),過(guò)A,C,D三點(diǎn)的☉O交AB于點(diǎn)E,已知BD=AD,∠BAD=2∠DAC=36°.(1)求證:AD是☉O的直徑;(2)過(guò)點(diǎn)E作EF⊥BC于點(diǎn)F,求證:EF與☉O相切.類型三作垂直,證半徑7.如圖,O為正方形ABCD對(duì)角線AC上一點(diǎn),以O(shè)為圓心,OA長(zhǎng)為半徑的☉O與BC相切于點(diǎn)M.求證:CD與☉O相切.8.如圖,AB是☉O的直徑,AM,BN分別切☉O于點(diǎn)A,B,CD交AM,BN于點(diǎn)D,C,DO平分∠ADC.(1)求證:CD是☉O的切線;(2)若AD=4,BC=9,求OD的長(zhǎng).答案全解全析1.證明∵AB=AC,∴∠C=∠B.∵OD∥BC,∴∠C=∠ODA=45°,∴∠B=45°,∴∠CAB=180°-45°-45°=90°,即AC⊥AB.∴AC是☉O的切線.2.證明∵AB是☉O的直徑,∴∠ADB=90°,∴∠BAD+∠B=90°.∵OB=OD,∴∠B=∠ODB,∵∠ODB=∠CDE,∠CDE=∠CAD,∴∠B=∠CAD,∴∠BAC=∠BAD+∠CAD=∠BAD+∠B=90°,即BA⊥AC,∴AC與☉O相切.3.證明如圖,連接OC,∵AC平分∠DAB,∴∠DAC=∠BAC.∵OC=OA,∴∠BAC=∠ACO,∴∠DAC=∠ACO,∴OC∥AD.∵CD⊥AD,∴OC⊥DC,∴CD是☉O的切線.4.解析(1)證明:如圖,連接OC,∵AB為直徑,∴∠ACB=90°,即∠BCO+∠OCA=90°.又∵∠DCB=∠CAD,∠CAD=∠OCA,∴∠OCA=∠DCB,∴∠DCB+∠BCO=90°,即∠DCO=90°.∵OC是☉O的半徑,∴CD是☉O的切線.(2)∵∠DCO=90°,∴OC2+CD2=OD2,∵OC=OB,CD=4,DB=2,∴OB2+42=(OB+2)2,∴OB=3,∴AB=6.∵AE⊥AD,AB是☉O的直徑,∴AE是☉O的切線.∵CD是☉O的切線,∴AE=CE,∵AD2+AE2=DE2,∴(6+2)2+AE2=(4+AE)2,解得AE=6.5.證明(1)∵AB=AC,∴∠ABC=∠ACB.又∵∠ACB=∠BCD,∠ABC=∠ADC,∴∠BCD=∠ADC,∴ED=EC.(2)如圖,連接OA,∵AB=AC,∴AB=AC,∴OA⊥BC.∵CA=CF,∴∠CAF=∠CFA,∴∠ACD=∠CAF+∠CFA=2∠CAF.∵∠ACB=∠BCD,∴∠ACD=2∠ACB,∴∠CAF=∠ACB,∴AF∥BC,∵OA⊥BC,∴OA⊥AF,∴AF為☉O的切線.6.證明(1)∵BD=AD,∴∠B=∠BAD=36°,∴∠ADC=72°.∵∠BAD=2∠DAC=36°,∴∠DAC=12∴∠ADC+∠DAC=90°,∴∠C=90°,∴AD是☉O的直徑.(2)如圖,連接OE,∵EF⊥BC,∴∠EFC=90°.∵OE=OA,∴∠OEA=∠BAD=36°,∴∠OEA=∠B,∴OE∥BC,又EF⊥BC,∴OE⊥EF,∴EF與☉O相切.7.證明如圖,連接OM,過(guò)點(diǎn)O作ON⊥CD于點(diǎn)N.∵☉O與BC相切于點(diǎn)M,∴OM⊥BC.∵O為正方形ABCD對(duì)角線AC上一點(diǎn),∴CO平分∠BCD.∴OM=ON,∴CD與☉O相切.8.解析(1)證明:如圖,過(guò)O點(diǎn)作OE⊥CD于點(diǎn)E,∵AM切☉O于點(diǎn)A,∴OA⊥AD.∵DO平分∠ADC,∴OE=OA.∵OA為☉O的半徑,∴OE是☉O的半徑,又OE⊥DC,∴CD是☉O的切線.(2)如圖,過(guò)D作DF⊥BC于F,∵AB是☉O的直徑,AM,BN分別切☉O于點(diǎn)A,B,∴AB⊥AD,AB⊥BC,∴四邊形ABFD為矩形,∴BF=AD=4,∴