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Last edited by isee 2021-8-14 00:22在三角形`ABC`中,点`P`在`AB`上,点`Q`在`AC`上,若`AP=\dfrac 25 AB`,`AQ=\dfrac 34 AC`,若点`N`为`BC`中点且中线`AN`交`PQ`于`M`,则`AN/AM=`_____.
PS:剪切板的图片也能直接上传,哈哈哈哈哈,才发现
解:用向量定比分点公式即可
\begin{align*}
\vv{AN}&=\frac 12\left(\vv{AB}+\vv{AC}\right)\\[1em]
&=\frac 12\left(\frac 52\vv{AP}+\frac 43\vv{AQ}\right)\\[1em]
&=\frac 54\vv{AP}+\frac 23\vv{AQ}\\[1em]
&=\frac {23}{12}\vv{AM}
\end{align*} |
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