# 逻辑通路/孟氏定理

D、E、F 三点共线 ${\displaystyle \Longleftrightarrow \left({\frac {\overrightarrow {BD}}{\overrightarrow {DC}}}\right)\left({\frac {\overrightarrow {CE}}{\overrightarrow {EA}}}\right)\left({\frac {\overrightarrow {AF}}{\overrightarrow {FB}}}\right)=-1}$

### 证明

• 我们先证明如果 D、E、F 三点共线的话，则上面所提的三个“有向比”的乘积为 -1

${\displaystyle {\dfrac {\overrightarrow {\mbox{BD}}}{\overrightarrow {\mbox{DC}}}}={\dfrac {\overrightarrow {\mbox{BH}}}{\overrightarrow {\mbox{IC}}}},\quad {\dfrac {\overrightarrow {\mbox{CE}}}{\overrightarrow {\mbox{EA}}}}={\dfrac {\overrightarrow {\mbox{CI}}}{\overrightarrow {\mbox{GA}}}},\quad {\dfrac {\overrightarrow {\mbox{AF}}}{\overrightarrow {\mbox{FB}}}}={\dfrac {\overrightarrow {\mbox{AG}}}{\overrightarrow {\mbox{HB}}}}}$

 ${\displaystyle \left({\frac {\overrightarrow {BD}}{\overrightarrow {DC}}}\right)\left({\frac {\overrightarrow {CE}}{\overrightarrow {EA}}}\right)\left({\frac {\overrightarrow {AF}}{\overrightarrow {FB}}}\right)}$ = ${\displaystyle \left({\frac {\overrightarrow {BH}}{\overrightarrow {IC}}}\right)\left({\frac {\overrightarrow {CI}}{\overrightarrow {GA}}}\right)\left({\frac {\overrightarrow {AG}}{\overrightarrow {HB}}}\right)}$ = ${\displaystyle \left({\frac {\overrightarrow {CI}}{\overrightarrow {IC}}}\right)\left({\frac {\overrightarrow {AG}}{\overrightarrow {GA}}}\right)\left({\frac {\overrightarrow {BH}}{\overrightarrow {HB}}}\right)}$ .mw-parser-output .nowrap,.mw-parser-output .nowrap a:before,.mw-parser-output .nowrap .selflink:before{white-space:nowrap}    ..... 根据“有向比性质 (2)” = (-1)(-1)(-1) = -1

• 其次，我们来证明：如果三个“有向比”的乘积为 -1，则 D、E、F 三点共线

(1) 直线 ${\displaystyle {\overset {\longleftrightarrow }{\scriptstyle {\text{DE}}}}}$ 与直线 ${\displaystyle {\overset {\longleftrightarrow }{\scriptstyle {\text{BC}}}}}$ 平行
(2) 直线 ${\displaystyle {\overset {\longleftrightarrow }{\scriptstyle {\text{DE}}}}}$ 与直线 ${\displaystyle {\overset {\longleftrightarrow }{\scriptstyle {\text{BC}}}}}$ 相交