專題六圓的相關(guān)證明與計(jì)算1.(原創(chuàng))如圖,AB為☉O的弦,M為☉O外一點(diǎn),且MB是☉O的切線,OM⊥AB于點(diǎn)C,交☉O于點(diǎn)D,P為AB上一點(diǎn),連接AP,BP,OA.(1)求證:∠P-∠M=90°.(2)若OA=6,tanM=342.(2024·福州模擬)如圖,AB是☉O的直徑,BD是☉O的弦,P是☉O外的一點(diǎn),PC⊥AB,垂足為C,PC與BD相交于點(diǎn)E,連接PD,AD,且∠PDB=∠DAB,延長PD交BA的延長線于點(diǎn)F.(1)求證:PD=PE.(2)若DF=4,PE=72,cos∠PFC=43.如圖,△ABC內(nèi)接于☉O,AB=AC,∠BAC=α,E是☉O上一點(diǎn),點(diǎn)C關(guān)于直線AE的對(duì)稱點(diǎn)為點(diǎn)D,連接AD,BD,BD交AC于點(diǎn)F.備用圖(1)已知∠CBD=20°,求∠BAD的度數(shù)(用含α的式子表示).(2)求證:BD經(jīng)過點(diǎn)E.(3)若E是AC的中點(diǎn),求證:DEEF=BC4.(2024·泉州三模)如圖,AD為☉O的直徑,點(diǎn)B、C在☉O上,AD⊥BC于點(diǎn)P.連接BO并延長交AC于點(diǎn)F.(1)若∠BAD=30°,試求∠C的度數(shù).(2)過點(diǎn)D作DE∥BF交☉O于點(diǎn)E,H為DE的中點(diǎn),連接OH.①求證:BP=OH;②連接CE,若DE=DP,△AOF的面積為92參考答案1.解析:(1)證明:如圖,連接OB,在優(yōu)弧AB上取一點(diǎn)E,連接AE,BE,則有∠E+∠P=180°.∵OM⊥AB,垂足為C,∴∠E=12∵M(jìn)B是☉O的切線,∴∠OBM=90°,∴∠BOM+∠M=90°,∴∠E+∠M=90°,∴180°-∠P+∠M=90°,∴∠P-∠M=90°.(2)∵∠OBM=∠OCB=90°,∴∠OBC=∠M.∵tanM=34∴在Rt△OBC中,tan∠OBC=OCBC=3∵OA=6,∴OB=OA=6,∴sin∠OBC=OCOB=OC6=∴OC=185,∴CD=OD-OC=6-185=2.解析:(1)證明:∵AB是☉O的直徑,∴∠ADB=90°,∴∠B+∠DAB=90°.∵PC⊥AB,∴∠PCB=90°,∴∠B+∠BEC=90°,∴∠BEC=∠DAB.又∵∠PDB=∠DAB,∠BEC=∠PED,∴∠PDE=∠PED,∴PD=PE.(2)∵PD=PE,PE=72∴PD=72∵DF=4,∴PF=PD+DF=152∵cos∠PFC=45∴CF=PF·cos∠PFC=152×4如圖,連接OD.由(1)可得∠B+∠PDE=∠B+∠BAD=90°,∵OB=OD,∴∠B=∠ODB,∴∠ODB+∠PDE=90°,∴∠ODF=90°,∴OF=DFcos∠PFC=4∴OC=CF-OF=6-5=1,根據(jù)勾股定理可得OD=OF2-D∴OA=OD=3,∴AC=OA+OC=3+1=4.3.解析:(1)∵CE=CE,∴∠CAE=∠CBD=20°.∵點(diǎn)C與點(diǎn)D關(guān)于AE對(duì)稱,∴∠DAE=∠CAE=20°,∴∠BAD=∠BAC+∠CAE+∠DAE=α+40°.(2)證明:∵點(diǎn)C與點(diǎn)D關(guān)于AE對(duì)稱,∴∠AED=∠AEC.∵四邊形ABCE內(nèi)接于☉O,∴∠AEC+∠ABC=180°.∵AB=AC,∴∠ABC=∠ACB.∵AB=AB,∴∠ACB=∠AEB,∴∠AEB+∠AED=180°,∴點(diǎn)B,D,E共線,∴BD過點(diǎn)E.(3)證明:如圖.∵E是AC的中點(diǎn),∴AE=CE,∴∠ABE=∠CBE=∠CAE,AE=CE.∵AB=AB,∴∠AEB=∠ACB,∴△BCF∽△BEA,∴BCBE=CFAE,∴BCCF∵∠AEF=∠AEB,∠FAE=∠ABE,∴△AEF∽△BEA,∴BEAE=AE∴BCCF=CE∵DE=CE,∴DEEF=BC4.解析:(1)∵AD為☉O的直徑,AD⊥BC,∴AD垂直平分BC,∴∠BAC=2∠BAD=60°,AB=AC,∴△ABC為等邊三角形,∴∠C=60°.(2)①證明:如圖1,連接AE.∵AD是直徑,∴∠AED=90°.∵H是DE的中點(diǎn),AO=DO,∴OH是△ADE的中位線,∴OH∥AE,∴∠OHD=∠AED=90°.∵DE∥BF,∴∠BOP=∠ODH.又∵∠OPB=∠OHD=90°,BO=DO,∴△BOP≌△ODH(AAS),∴BP=OH.②如圖2,過點(diǎn)F作FN⊥AD,交AD于點(diǎn)N,設(shè)DE=DP=2x,則DH=HE=x.∵△BOP≌△ODH,∴OP=DH=x,∴OD=3x=OA=OB,∴BP=OB2-O∵AD⊥BC,FN⊥AD,∴tan∠BAP=BPAP,tan∠CAP=NF∵∠BAP=∠CAP,∴BPAP=NF∴NFAN=2∴AN=2NF.∵∠BOP=∠NOF,∴tan∠BOP=BPOP,ta