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\[\displaystyle \frac{BP}{PC} \cdot \frac{CQ}{QA} \cdot \frac{AR}{RB}=1\]
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\[\displaystyle \frac{\triangle{OAB}}{\triangle{OCA}}=\frac{BP}{PC}\]
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\(\displaystyle \triangle{OAB}=AO \times BH \times \frac{1}{2}\)ââ¡
\(\displaystyle \triangle{OAC}=AO \times CK \times \frac{1}{2}\)ââ¢
â¡,â¢ã§AOãš\(\frac{1}{2}\)ãæ¶å»ãããšã
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\[\displaystyle \frac{\triangle{OAB}}{\triangle{OCA}}=\frac{BP}{PC}\]
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\(\displaystyle \frac{\triangle{OAB}}{\triangle{OCA}}=\frac{CQ}{QA}\)ââŠ
\(\displaystyle \frac{\triangle{OBC}}{\triangle{OAB}}=\frac{AR}{RB}\)ââ§
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\begin{eqnarray}
&&\frac{BP}{PC} \cdot \frac{CQ}{QA} \cdot \frac{AR}{RB}\\
&=&\frac{\triangle{OAB}}{\triangle{OCA}} \cdot \frac{\triangle{OBC}}{\triangle{OAB}} \cdot \frac{\triangle{OCA}}{\triangle{ABC}}\\
&=&1
\end{eqnarray}
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\begin{eqnarray}
\displaystyle \frac{QC}{AQ} \cdot \frac{PB}{CP} \cdot \frac{RA}{BR}&=&1\\
\displaystyle \frac{3}{2} \cdot \frac{1}{3} \cdot \frac{AR}{2}&ïŒ&1\\
AR&ïŒ&4
\end{eqnarray}
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\[\frac{2}{AR} \cdot \frac{3}{1} \cdot \frac{2}{3}ïŒ1\]
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\displaystyle \frac{QC}{AQ} \cdot \frac{PB}{CP} \cdot \frac{RA}{BR}&ïŒ&1\\
\displaystyle \frac{3}{1} \cdot \frac{1}{1} \cdot \frac{RA}{BR}&ïŒ&1\\
\displaystyle \frac{RA}{BR}&ïŒ&3
\end{eqnarray}
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\begin{eqnarray}
\displaystyle \frac{PN}{BP} \cdot \frac{CA}{NC} \cdot \frac{MB}{AM}&ïŒ&1\\
\displaystyle \frac{PN}{BP} \cdot \frac{7}{3} \cdot \frac{5}{2}&ïŒ&1\\
\displaystyle \frac{PN}{BP}&=&\frac{6}{35}
\end{eqnarray}
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\begin{eqnarray}
\displaystyle \frac{NC}{AN} \cdot \frac{QB}{CQ} \cdot \frac{MA}{BM}&=&1\\
\displaystyle \frac{3}{4} \cdot \frac{QB}{CQ} \cdot \frac{2}{5}&=&1\\
\displaystyle \frac{QB}{CQ}&=&\frac{10}{3}
\end{eqnarray}
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