AB→=a⃗\overrightarrow{AB} = \vec{a}AB=a 、 AC→=b⃗\overrightarrow{AC} = \vec{b}AC=b 、 AD→=c⃗\overrightarrow{AD} = \vec{c}AD=c とおく {CD→=−b⃗+c⃗BC→=−a⃗+b⃗BD→=−a⃗+c⃗\begin{cases} \overrightarrow{CD} = -\vec{b} + \vec{c} \\ \overrightarrow{BC} = -\vec{a} + \vec{b} \\ \overrightarrow{BD} = -\vec{a} + \vec{c} \end{cases}⎩⎨⎧CD=−b+cBC=−a+bBD=−a+c よって AB2+CD2=AD2+BC2AB^2 + CD^2 = AD^2 + BC^2AB2+CD2=AD2+BC2 (a⃗)2+(−b⃗+c⃗)2=(c⃗)2+(−a⃗+b⃗)2(\vec{a})^2 + (-\vec{b}+\vec{c})^2 = (\vec{c})^2 + (-\vec{a}+\vec{b})^2(a)2+(−b+c)2=(c)2+(−a+b)2 a2+b2−2b⃗c⃗+c2=c2+a2−2a⃗b⃗+b2a^2 + b^2 - 2\vec{b}\vec{c} + c^2 = c^2 + a^2 - 2\vec{a}\vec{b} + b^2a2+b2−2bc+c2=c2+a2−2ab+b2 b⃗c⃗−a⃗b⃗=0\vec{b}\vec{c} - \vec{a}\vec{b} = 0bc−ab=0 b⃗(−a⃗+c⃗)=0\vec{b}(-\vec{a}+\vec{c}) = 0b(−a+c)=0 AC→⋅BD→=0\overrightarrow{AC} \cdot \overrightarrow{BD} = 0AC⋅BD=0 ∴AC⊥BD\therefore AC \perp BD∴AC⊥BD