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Generalization of Ptolemy's theorem, Tran Quang Hung
设ABCD为凸四边形。角度的绝对差\begin{array}{l}\psi=|\angle A D B-\angle A C B|=|\angle D A C-\angle D B C| \\ \phi=|\angle A B D-\angle A C D|=|\angle B A C-\angle B D C|\end{array}则$A C \cdot B D=A D \cdot B C \cdot \cos \psi+A B \cdot C D \cdot \cos \phi$
证明
设$P$使$\triangle APD\sim\triangle ABC,\triangle APB\sim\triangle ADC$. 根据线段比
$$A D \cdot B C=P D \cdot A C \Rightarrow P D=\frac{A D \cdot B C}{A C}$$
与$$A B \cdot C D=P B \cdot A C \Rightarrow P B=\frac{A B \cdot C D}{A C}$$注意到$∠PBD=|∠PBA-∠ABD|=|∠ACD-∠ABD|=ϕ$ |
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