梅涅劳斯定理

如情况一，连接${\displaystyle AL}$、${\displaystyle CN}$，有

${\displaystyle {\frac {AN}{NB}}\cdot {\frac {BL}{LC}}\cdot {\frac {CM}{MA}}={\frac {\mathrm {S} _{\triangle LNA}}{\mathrm {S} _{\triangle LNB}}}\cdot {\frac {\mathrm {S} _{\triangle LNB}}{\mathrm {S} _{\triangle LNC}}}\cdot {\frac {\mathrm {S} _{\triangle LNC}}{\mathrm {S} _{\triangle ANL}}}=1}$

如情況一，设${\displaystyle \angle ANM=\alpha }$，${\displaystyle \angle AMN=\beta }$，${\displaystyle \angle MLC=\gamma }$，則在${\displaystyle \triangle AMN}$中由正弦定理，有

${\displaystyle {\frac {AN}{AM}}={\frac {\sin \beta }{\sin \alpha }},}$

同理，因對頂角相等在${\displaystyle \triangle NBL}$和${\displaystyle \triangle CLM}$中有

${\displaystyle {\frac {BL}{BN}}={\frac {\sin \alpha }{\sin \gamma }},}$

及

${\displaystyle {\frac {CM}{CL}}={\frac {\sin \gamma }{\sin \beta }}.}$

三式相乘即得

${\displaystyle {\frac {AN}{AM}}\cdot {\frac {BL}{BN}}\cdot {\frac {CM}{CL}}={\frac {\sin \beta }{\sin \alpha }}\cdot {\frac {\sin \alpha }{\sin \gamma }}\cdot {\frac {\sin \gamma }{\sin \beta }}=1,}$

即

${\displaystyle {\frac {AN}{NB}}\cdot {\frac {BL}{LC}}\cdot {\frac {CM}{MA}}=1.}$