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信息發(fā)布者：
證明：????$(1)$????∵????$AD//BC$????
∴????$∠ACF=∠DAC$????
∵????$∠FAC=∠ADE，$????????$AC=AD$????
∴????$△ACF≌△DAE(\mathrm {ASA})$????
∴????$AF=DE$????
????$(2)$????∵????$△ACF≌△DAE$????
∴????$∠AFC=∠DEA$????
∴????$∠AFB=∠DEC$????
∵????$∠ABC=∠CDE$????
∴????$△ABF∽△CDE$????
∴????$\frac {AF}{CE}=\frac {BF}{DE}$????
∴????$AF·DE=BF·CE$????
∵????$AF=DE$????
∴????$AF^2=BF·CE$????
????$(1)$????證明：∵????$AG⊥BC，$????????$AF⊥DE$????
∴????$∠AFE=∠AGC=90°$????
∵????$∠EAF=∠GAC$????
∴????$∠AED=∠ACB$????
∵????$∠EAD=∠CAB$????
∴????$△ADE∽△ABC $????
????$(2) $????由????$(1)$????可知????$△ADE∽△ABC$????
∴????$\frac {AE}{AC}=\frac {AD}{AB}=\frac {3}{5}$????
由????$(1)$????可知????$∠AFE=∠AGC=90°$????
∵????$∠EAF=∠GAC$????
∴????$△EAF∽△CAG$????
∴????$\frac {AF}{AG}=\frac {AE}{AC}$????
∴????$AF：$????????$AG=\frac {3}{5}$????
證明：????$(1) $????連接????$OC$????
∵????$AC$????平分????$∠DAB$????
∴????$∠DAC=∠EAC$????
∵????$OA=OC$????
∴????$∠OCA=∠EAC$????
∴????$∠DAC=∠OCA$????
∴????$OC//AD$????
∵????$CD⊥AD$????
∴????$OC⊥CD$????
∵????$OC$????是????$⊙O$????的半徑
∴????$CD$????是????$⊙O$????的切線 
????$(2)$????∵????$AC$????平分????$∠DAB$????
∴????$∠DAC=∠EAC$????
又∵????$CD⊥AD，$????????$FG⊥AB$????
∴????$∠AGF=∠D=90°$????
∴????$∠AFG+∠DAG=90°，$????????$∠E+∠DAE=90°$????
∴????$∠AFG=∠E$????
∴????$△AHF∽△ACE$????
∴????$\frac {AH}{AC}=\frac {AF}{AE}，$????即????$AF·AC=AE·AH$????
解：????$(1) $????由題意，可得????$∠AOB=90°，$????????$OA=8，$????????$OB=6$????
∴????$AB=10$????
∵????$C$????是線段????$AB$????的中點(diǎn)
∴????$AC=5$????
①如圖①，如果點(diǎn)????$P $????與點(diǎn)????$B$????對(duì)應(yīng)，那么????$△PAC∽△BAO$????
∴????$\frac {PA}{BA}=\frac {AC}{AO}$????
∴????$PA=\frac {25}{4}$????
∴????$OP=OA-PA=\frac {7}{4}$????
∴????$P(\frac {7}{4} ，$????????$0)$????
 ②如圖②，如果點(diǎn)????$P $????與點(diǎn)????$O$????對(duì)應(yīng)，那么????$△PAC∽△OAB$????
∴????$\frac {PA}{OA}=\frac {AC}{AB}$????
∴????$PA= 4$????
∴????$OP=OA-PA=4$????
∴????$P(4，$????????$0)$????
綜上所述，點(diǎn)????$P$????的坐標(biāo)為????$(\frac {7}{4}，$????????$0 ) $????或????$(4，$??????
??$0)$??

?
???$(2)M(\frac {8}{3} ，$????????$0)$????
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