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第8頁

信息發(fā)布者：
證明：?$(1)$?連接?$OA，$?如圖：


?$∵AB⊥CD，$?
?$∴∠AFD=90°，$?
?$∴∠FAD+∠ADF=90°，$?
?$∵OA=OD，$?
?$∴∠OAD=∠ADF，$?
?$∴∠FAD+∠OAD=90°，$?
?$∵∠EAD=∠FAD，$?
?$∴∠EAD+∠OAD=90°，$?即?$∠OAE=90°，$?
?$∴OA⊥AE，$?
?$∵OA$?是?$⊙O$?半徑，
?$∴AE$?是?$⊙O$?的切線.
?$(2)$?解：連接?$AC，$??$AO，$?如圖：


?$∵CD$?為?$⊙O$?直徑，
?$∴∠CAD=90°，$?
?$∴∠C+∠ADC=90°，$?
?$∵∠FAD+∠ADC=90°，$?
?$∴∠C=∠FAD，$?
?$∵∠EAD=∠FAD，$?
?$∴∠C=∠EAD，$?
?$∵∠P=∠P，$?
?$∴△ADP∽△CAP，$?
?$∴\frac {AP}{CP}=\frac {PD}{AP}，$?
?$∵PA=4，$??$PD=2，$?
?$∴\frac {4}{CP}=\frac {2}{4}，$?
解得?$CP=8，$?
?$∴CD=CP-PD=8-2=6，$?
?$∴⊙O$?的半徑為?$3.$?
?$∴OA=3=OD，$?
?$∴OP=OD+PD=5，$?
?$∵∠OAP=90°=∠DEP，$??$∠P=∠P，$?
?$∴△OAP∽△DEP，$?
?$∴\frac {DE}{OA}=\frac {PD}{OP}，$?即?$\frac {DE}{3}=\frac {2}{5}，$?
?$∴DE=\frac {6}{5}，$?
?$∴⊙O$?的半徑為?$3，$??$DE$?的長為?$\frac {6}{5}.$?

?$\frac {1}{2}$?或?$2$?或?$\frac {14}{5} $?
解：?$(1)∵∠COB=90°，$??$∠CBO=45°，$??$B(3，$??$0)，$?
?$∴∠OCB=45°=∠CBO，$?
?$∴OC=OB=3，$?
?$∵CD∥OA，$??$∠D=90°，$??$∠COA=90°，$?
?$∴∠DCO=90°，$??$∠DAO=90°，$?
∴四邊形?$COAD$?是矩形，
?$∵A(5，$??$0)，$?
?$∴CD=OA=5，$??$OC=AD=3，$?
∴點?$D$?的坐標(biāo)為?$(5，$??$3)，$?
∴點?$C$?的坐標(biāo)為?$(0，$??$3).$?
?$(2)$?如圖，當(dāng)?$P$?在?$B$?的左側(cè)，
?$∵∠BCP=15°，$??$∠OCB=45°，$?
?$∴∠PCO=30°，$?
?$OP=OC×tan_{30}°=3×\frac {\sqrt{3}}{3}=\sqrt{3}，$?
?$∵Q(-4，$??$0)，$?
?$∴QP=4+\sqrt{3}，$?
即?$t=(4+\sqrt{3})÷2=\frac {4+\sqrt{3}}{2}；$?
②如圖，當(dāng)?$P$?在?$B$?的右側(cè)，
?$∵∠BCP=15°，$??$∠OCB=45°，$?
?$∴∠OCP=60°，$?
則?$∠CPO=30°，$?
?$∴OP=\sqrt{3}OC=3\sqrt{3}，$?
?$∴QP=4+3\sqrt{3}，$?
?$t=(4+3\sqrt{3})÷2=\frac {4+3\sqrt{3}}{2}.$?
綜上可知，當(dāng)?$∠BCP=15°$?時，?$t$?的值為?$\frac {4+\sqrt{3}}{2}$?或?$\frac {4+3\sqrt{3}}{2}.$?
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