托勒密定理(Ptolemy's theorem):圓內接四邊形對角線之積等於兩對邊相乘之和

圓內接四邊形\(ABCD\)，則\(\overline{AC}\times \overline{BD} = \overline{AB}\times \overline{CD} +\overline{BC}\times \overline{AD}\)

$$\cases{\triangle ABD \Rightarrow \cos  \alpha={a^2+d^2-f^2\over 2ad} \\\triangle BCD \Rightarrow \cos (\pi-\alpha)={b^2+c^2-f^2\over 2bc}}  \Rightarrow {a^2+d^2-f^2\over 2ad} =-{b^2+c^2-f^2\over 2bc} \\ \Rightarrow {a^2+d^2\over ad}-{1\over ad}f^2 ={1\over bc}f^2-{b^2+c^2\over bc} \Rightarrow f^2\\={(a^2+d^2)bc+(b^2+c^2)ad \over ad+bc}={(ab+cd)(ac+bd)\over ad+bc}\cdots (1)$$

$$同理\cases{\triangle ABC \Rightarrow \cos  \beta={a^2+b^2-e^2\over 2ab} \\\triangle ACD \Rightarrow \cos (\pi-\beta)={c^2+d^2-e^2\over 2cd}}  \Rightarrow {a^2+b^2-e^2\over 2ab} =-{c^2+d^2-e^2\over 2cd} \\ \Rightarrow {a^2+b^2\over ab}-{1\over ab}e^2 ={1\over cd}e^2-{c^2+d^2\over cd} \Rightarrow e^2\\={(a^2+b^2)cd+(c^2+d^2)ab \over ab+cd} ={(ac+bd)(ad+bc) \over ab+cd}\cdots (2)\\ 因此(1)\times (2) \Rightarrow e^2f^2 ={(ac+bd)(ad+bc) \over ab+cd} \times{(ab+cd)(ac+bd) \over ad+bc} =(ac+bd)^2\\ \Rightarrow ef=ac+bd，故得證$$