塞瓦定理

${\displaystyle \because \quad {\frac {\overline {BD}}{\overline {DC}}}={\frac {\mathrm {S} _{\triangle ABD}}{\mathrm {S} _{\triangle ADC}}}={\frac {\mathrm {S} _{\triangle OBD}}{\mathrm {S} _{\triangle ODC}}}.}$

由等比性质,

${\displaystyle {\frac {\overline {BD}}{\overline {DC}}}={\frac {\mathrm {S} _{\triangle ABD}-\mathrm {S} _{\triangle OBD}}{\mathrm {S} _{\triangle ADC}-\mathrm {S} _{\triangle ODC}}}={\frac {\mathrm {S} _{\triangle ABO}}{\mathrm {S} _{\triangle CAO}}}.}$

同理 ${\displaystyle {\frac {\overline {CE}}{\overline {EA}}}={\frac {\mathrm {S} _{\triangle BCO}}{\mathrm {S} _{\triangle ABO}}},\;{\frac {\overline {AF}}{\overline {FB}}}={\frac {\mathrm {S} _{\triangle CAO}}{\mathrm {S} _{\triangle BCO}}}.}$

${\displaystyle \therefore \quad {\frac {\overline {BD}}{\overline {DC}}}\cdot {\frac {\overline {CE}}{\overline {EA}}}\cdot {\frac {\overline {AF}}{\overline {FB}}}={\frac {\mathrm {S} _{\triangle ABO}}{\mathrm {S} _{\triangle CAO}}}\cdot {\frac {\mathrm {S} _{\triangle BCO}}{\mathrm {S} _{\triangle ABO}}}\cdot {\frac {\mathrm {S} _{\triangle CAO}}{\mathrm {S} _{\triangle BCO}}}=1.}$

证毕。[1][2]

在三角形${\displaystyle ABC}$中，${\displaystyle \angle A}$的角平分線交${\displaystyle {\overline {BC}}}$於${\displaystyle D}$，${\displaystyle {\frac {\overline {DB}}{\overline {DC}}}={\frac {\overline {AB}}{\overline {AC}}}}$。

• 梅涅劳斯定理
• 莫雷角三分線定理

1. Russell, John Wellesley. . . Clarendon Press. 1905.
2. Alfred S. Posamentier and Charles T. Salkind (1996), Challenging Problems in Geometry, pages 177–180, Dover Publishing Co., second revised edition.