トレミーの定理
トレミーの定理
4角形ABCDが円に内接してるとき、
\[ \left|\overrightarrow{AB}\right|\left|\overrightarrow{CD}\right|+\left|\overrightarrow{BC}\right|\left|\overrightarrow{DA}\right|=\left|\overrightarrow{BD}\right|\left|\overrightarrow{CA}\right| \] が成り立つ。
4角形ABCDが円に内接してるとき、
\[ \left|\overrightarrow{AB}\right|\left|\overrightarrow{CD}\right|+\left|\overrightarrow{BC}\right|\left|\overrightarrow{DA}\right|=\left|\overrightarrow{BD}\right|\left|\overrightarrow{CA}\right| \] が成り立つ。
(0)
4角形ABCDが円に内接しているので、対頂角の和は180°になるので\(\cos\left(A+C\right)=-1\)となる。オイラーの定理より、
\begin{align*} \left|\overrightarrow{BD}\right|^{2}\left|\overrightarrow{CA}\right|^{2} & =\left|\overrightarrow{AB}\right|^{2}\left|\overrightarrow{CD}\right|^{2}+\left|\overrightarrow{BC}\right|^{2}\left|\overrightarrow{DA}\right|^{2}-2\left|\overrightarrow{AB}\right|\left|\overrightarrow{BC}\right|\left|\overrightarrow{CD}\right|\left|\overrightarrow{DA}\right|\cos\left(A+C\right)\\ & =\left|\overrightarrow{AB}\right|^{2}\left|\overrightarrow{CD}\right|^{2}+\left|\overrightarrow{BC}\right|^{2}\left|\overrightarrow{DA}\right|^{2}+2\left|\overrightarrow{AB}\right|\left|\overrightarrow{BC}\right|\left|\overrightarrow{CD}\right|\left|\overrightarrow{DA}\right|\\ & =\left(\left|\overrightarrow{AB}\right|\left|\overrightarrow{CD}\right|+\left|\overrightarrow{BC}\right|\left|\overrightarrow{DA}\right|\right)^{2} \end{align*} これより、辺の長さは正なので、
\begin{align*} \left|\overrightarrow{BD}\right|\left|\overrightarrow{CA}\right| & =\left|\overrightarrow{AB}\right|\left|\overrightarrow{CD}\right|+\left|\overrightarrow{BC}\right|\left|\overrightarrow{DA}\right| \end{align*} となる。
(0)-2
\begin{align*} \left|\overrightarrow{AB}\right|\left|\overrightarrow{CD}\right|+\left|\overrightarrow{BC}\right|\left|\overrightarrow{DA}\right| & =\left|\overrightarrow{AB}\right|\left(\frac{\overrightarrow{CD}\cdot\overrightarrow{CD}}{\left|\overrightarrow{CD}\right|}\right)+\left|\overrightarrow{BC}\right|\left(\frac{\overrightarrow{DA}\cdot\overrightarrow{DA}}{\left|\overrightarrow{DA}\right|}\right)\\ & =\left|\overrightarrow{AB}\right|\left(\frac{\overrightarrow{CD}\cdot\left(\overrightarrow{CB}+\overrightarrow{BD}\right)}{\left|\overrightarrow{CD}\right|}\right)+\left|\overrightarrow{BC}\right|\left(\frac{\overrightarrow{DA}\cdot\left(\overrightarrow{DB}+\overrightarrow{BA}\right)}{\left|\overrightarrow{DA}\right|}\right)\\ & =\left|\overrightarrow{AB}\right|\left(\left|\overrightarrow{CB}\right|\frac{\overrightarrow{CD}\cdot\overrightarrow{CB}}{\left|\overrightarrow{CD}\right|\left|\overrightarrow{CB}\right|}+\left|\overrightarrow{BD}\right|\frac{\overrightarrow{CD}\cdot\overrightarrow{BD}}{\left|\overrightarrow{CD}\right|\left|\overrightarrow{BD}\right|}\right)+\left|\overrightarrow{BC}\right|\left(\left|\overrightarrow{DB}\right|\frac{\overrightarrow{DA}\cdot\overrightarrow{DB}}{\left|\overrightarrow{DA}\right|\left|\overrightarrow{DB}\right|}+\left|\overrightarrow{BA}\right|\frac{\overrightarrow{DA}\cdot\overrightarrow{BA}}{\left|\overrightarrow{DA}\right|\left|\overrightarrow{BA}\right|}\right)\\ & =\left|\overrightarrow{AB}\right|\left(-\left|\overrightarrow{CB}\right|\frac{\overrightarrow{AD}\cdot\overrightarrow{AB}}{\left|\overrightarrow{AD}\right|\left|\overrightarrow{AB}\right|}+\left|\overrightarrow{BD}\right|\frac{\overrightarrow{CA}\cdot\overrightarrow{BA}}{\left|\overrightarrow{CA}\right|\left|\overrightarrow{BA}\right|}\right)+\left|\overrightarrow{BC}\right|\left(\left|\overrightarrow{DB}\right|\frac{\overrightarrow{CA}\cdot\overrightarrow{CB}}{\left|\overrightarrow{CA}\right|\left|\overrightarrow{CB}\right|}+\left|\overrightarrow{BA}\right|\frac{\overrightarrow{DA}\cdot\overrightarrow{BA}}{\left|\overrightarrow{DA}\right|\left|\overrightarrow{BA}\right|}\right)\\ & =-\left|\overrightarrow{CB}\right|\frac{\overrightarrow{AD}\cdot\overrightarrow{AB}}{\left|\overrightarrow{AD}\right|}+\left|\overrightarrow{BD}\right|\frac{\overrightarrow{CA}\cdot\overrightarrow{BA}}{\left|\overrightarrow{CA}\right|}+\left|\overrightarrow{DB}\right|\frac{\overrightarrow{CA}\cdot\overrightarrow{CB}}{\left|\overrightarrow{CA}\right|}+\left|\overrightarrow{BC}\right|\frac{\overrightarrow{DA}\cdot\overrightarrow{BA}}{\left|\overrightarrow{DA}\right|}\\ & =\left|\overrightarrow{BD}\right|\frac{\overrightarrow{CA}\cdot\left(\overrightarrow{CB}+\overrightarrow{BA}\right)}{\left|\overrightarrow{CA}\right|}\\ & =\left|\overrightarrow{BD}\right|\left|\overrightarrow{CA}\right| \end{align*}
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3点を通る円
\[
x^{2}+y^{2}-\frac{1}{x_{1}y_{2}+y_{1}x_{3}+x_{2}y_{3}-x_{1}y_{3}-y_{1}x_{2}-y_{2}x_{3}}\left(\begin{array}{ccc}
x & y & 1\end{array}\right)\left(\begin{array}{ccc}
y_{2}-y_{3} & y_{3}-y_{1} & y_{1}-y_{2}\\
x_{3}-x_{2} & x_{1}-x_{3} & x_{2}-x_{1}\\
x_{2}y_{3}-y_{2}x_{3} & y_{1}x_{3}-x_{1}y_{3} & x_{1}y_{2}-y_{1}x_{2}
\end{array}\right)\left(\begin{array}{c}
x_{1}^{\;2}+y_{1}^{\;2}\\
x_{2}^{\;2}+y_{2}^{\;2}\\
x_{3}^{\;2}+y_{3}^{\;2}
\end{array}\right)=0
\]
傍心円の半径
\[
r_{a}=\frac{S}{s-a}
\]
4角形が円に外接するときの対辺の和
\[
\left|\overrightarrow{AB}\right|+\left|\overrightarrow{CD}\right|=\left|\overrightarrow{BC}\right|+\left|\overrightarrow{DA}\right|
\]
重心は中線を2:1に内分