# 托勒密定理

## 证明

### 几何证明

1. 设ABCD是圆内接四边形
2. BC上，圆周角∠BAC = ∠BDC，而在AB上，∠ADB = ∠ACB。
3. 在AC上取一点K，使得∠ABK = ∠CBD； 因为∠ABK + ∠CBK = ∠ABC = ∠CBD + ∠ABD，所以∠CBK = ∠ABD。
4. 因此△ABK与△DBC相似，同理也有△ABD相似于△KBC。
5. 因此AK/AB = CD/BD，且CK/BC = DA/BD；
6. 因此AK·BD = AB·CD，且CK·BD = BC·DA；
7. 两式相加，得(AK+CK)·BD = AB·CD + BC·DA；
8. 但AK+CK = AC，因此AC·BD = AB·CD + BC·DA。即得證

### 和差化积证明

${\displaystyle \sin(\alpha +\beta )\sin(\beta +\gamma )=\sin \alpha \sin \gamma +\sin \beta \sin(\alpha +\beta +\gamma )}$

### 逆定理的几何证明

${\displaystyle \angle ABP=\angle DBC}$ （红色角）

${\displaystyle \angle ABD=\angle PBC}$

${\displaystyle {\frac {AB}{DB}}={\frac {PB}{CB}}}$

${\displaystyle AB\cdot CD=AP\cdot BD}$
${\displaystyle AD\cdot BC=PC\cdot BD}$

${\displaystyle AB\cdot CD+AD\cdot BC=(AP+PC)\cdot BD\geqslant AC\cdot BD}$

${\displaystyle \angle ABC+\angle ADC=\angle ABC+\angle ADB+\angle BDC=\angle ABC+\angle PCB+\angle BAP}$
${\displaystyle =\angle ABC+\angle BAC+\angle BCA=\pi }$

### 反演的证明

${\displaystyle A'B'+B'C'=A'C'\qquad \qquad \qquad (*)}$

${\displaystyle A'B'=AB\cdot {\frac {DA'}{DB}}=AB\cdot {\frac {r^{2}}{DA\cdot DB}},\qquad \,B'C'=BC\cdot {\frac {r^{2}}{DC\cdot DB}},\qquad \,A'C'=AC\cdot {\frac {r^{2}}{DA\cdot DC}}}$

${\displaystyle AB\cdot {\frac {r^{2}}{DA\cdot DB}}+BC\cdot {\frac {r^{2}}{DC\cdot DB}}=AC\cdot {\frac {r^{2}}{DA\cdot DC}}}$

${\displaystyle AB\cdot CD+AD\cdot BC=AC\cdot BD}$

## 与西姆松定理的关系

${\displaystyle LN=PB\sin \angle LBN=PB\sin \angle ABC}$

${\displaystyle LN={\frac {PB\cdot AC}{2R}}}$

${\displaystyle NM={\frac {PA\cdot BC}{2R}},\qquad \quad LM={\frac {PC\cdot AB}{2R}}}$

${\displaystyle LN+NM\geqslant LM}$

${\displaystyle PA\cdot BC+PB\cdot AC\geqslant PC\cdot AB}$

## 推广

${\displaystyle AC^{2}\cdot BD^{2}=AB^{2}\cdot CD^{2}+BC^{2}\cdot AD^{2}-2AB\cdot BC\cdot CD\cdot DA\cdot \cos(\angle ABC+\angle ADC)}$ [6]

${\displaystyle AC^{2}\cdot BD^{2}=AB^{2}\cdot CD^{2}+BC^{2}\cdot AD^{2}-2AB\cdot BC\cdot CD\cdot DA\cdot \cos(\angle ABC+\angle ADC)}$
${\displaystyle .\quad \leqslant AB^{2}\cdot CD^{2}+BC^{2}\cdot AD^{2}+2AB\cdot BC\cdot CD\cdot DA\cdot }$

${\displaystyle (AC\cdot BD)^{2}\leqslant (AB\cdot CD+BC\cdot AD)^{2}}$
${\displaystyle AC\cdot BD\leqslant AB\cdot CD+BC\cdot AD}$

## 参考与注释

1. ^ R.A.约翰逊，《近代欧氏几何学》，第51页
2. ^ R.A.约翰逊，《近代欧氏几何学》，第52页
3. ^ R.A.约翰逊，《近代欧氏几何学》，第117页
4. ^ （英文）Harold Scott Macdonald Coxeter, Samuel L. Greitzer. Geometry revisited. The Mathematical Association of America; 1ST edition. 1967. ISBN 978-0883856192.,p.42
5. ^ R.A.约翰逊，《近代欧氏几何学》，第102-103页，原文如此
6. ^ R.A.约翰逊，《近代欧氏几何学》，第54页