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发表于 2026-8-21 11:53
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本帖最后由 dodonaomikiki 于 2026-8-21 12:38 编辑
推荐一种办法!供大咖们评鉴!
\(
Proof: \\
链接CE,CP,CQ,AE,AP, AQ \\
\Longrightarrow \displaystyle{ \boldsymbol { \frac{CM}{MA} = \frac{S \blacktriangle QCE}{ S \blacktriangle QAE } = \frac{CQ.CE.sin\angle QCE}{AQ.AE.sin\angle QAE} }} \\
\displaystyle{ \boldsymbol { =\frac{CQ.CP.sin(\angle ACP-\angle ACQ)}{AQ.AP.sin(\angle CAP-\angle CAQ)} \qquad \mathscr{A} }} \\
同理 \\
\displaystyle{ \boldsymbol { \frac{ AN}{ NB} = =\frac{AQ.AP.sin(\angle BAQ-\angle BAP)}{BQ.BP.sin(\angle ABP-\angle ABQ)} \qquad \mathscr{B} }} \\
\displaystyle{ \boldsymbol { \frac{BL}{LC } = =\frac{BQ.BP.sin(\angle CBQ+\angle CBP)}{CQ.CP.sin(\angle BCQ+ \angle BCP)} \qquad \mathscr{C} }} \\
容易晓得: \\
\displaystyle{ \boldsymbol { \angle ACP- \angle ACQ= \angle ABP- \angle ABQ }} \\
\displaystyle{ \boldsymbol { \angle CAP- \angle CAQ=180度-( \angle BCQ- \angle BCP) }} \\
\displaystyle{ \boldsymbol { \angle BAQ- \angle BAP=180度-( \angle BCQ+ \angle BCP) }} \\
最后我们得到: \displaystyle{ \boldsymbol { \mathscr{A} \times \mathscr{B} \times \mathscr{C} }} \\
\displaystyle{ \boldsymbol { = \frac{CM}{MA} \bullet \frac{ AN}{ NB} \bullet \frac{BL}{LC } = 1 }} \\
根据没涅劳斯定理
得到 L,M,N三点共线 \) |
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