本帖最后由 青青子衿 于 2019-4-27 23:09 编辑
回复 3# 郝酒
参见此贴第6楼
http://kuing.orzweb.net/redirect ... =4181&pid=23568
也可以参见此贴第8楼
http://kuing.orzweb.net/redirect ... =5433&pid=27165
\begin{align*}
\overrightarrow{AG}&=\dfrac{1}{3}\overrightarrow{AB}+\dfrac{1}{3}\overrightarrow{AC}\\
\\
\overrightarrow{AO}&=\dfrac{b\cos(B)\,}{a\cos(A)+b\cos(B)+c\cos(C)}\overrightarrow{AB}+\dfrac{c\cos(C)\,}{a\cos(A)+b\cos(B)+c\cos(C)}\overrightarrow{AC}\\
\\
\overrightarrow{AI}&=\dfrac{b}{a+b+c}\overrightarrow{AB}+\dfrac{c}{a+b+c}\overrightarrow{AC}\\
\overrightarrow{AH}&=\begin{split}
\,\\\,\\
\dfrac{b\cos(C)\cos(A)}{a\cos(B)\cos(C)+b\cos(C)\cos(A)+c\cos(A)\cos(B)}\overrightarrow{AB}\\
+\dfrac{c\cos(B)\cos(A)}{a\cos(B)\cos(C)+b\cos(C)\cos(A)+c\cos(A)\cos(B)}\overrightarrow{AC}
\end{split}\\
\,\\
\overrightarrow{AI_A}&=\dfrac{b}{-a+b+c}\overrightarrow{AB}+\dfrac{c}{-a+b+c}\overrightarrow{AC}\\
\end{align*} |