「数学」Menelaus定理与Ceva定理

内容

\(\rm Menelaus\)定理

已知三角形\(\triangle ABC\)被一直线所截,交三条边或三条边的延长线与点\(X, Y, Z\)点,则有

\[\frac{AX}{XB} \cdot \frac{BZ}{ZC} \cdot \frac{CY}{YA}=1 \]

(注:上图为一种情况,还有一种为“直线不经过三角形的任何一边,即与三角形的交点数为\(0\)”)

证明:

过点\(C\)\(CP // DF\)\(AB\)\(P\),则

\[\frac{BZ}{ZC}=\frac{BX}{XP}\tag{1} \]

\[\frac{CY}{YA}=\frac{PX}{XA}\tag{2} \]

\[(1) \times (2) \rm{得:} \frac{BZ}{ZC}\cdot \frac{CY}{YA}=\frac{BX}{XP}\cdot \frac{PX}{XA} \]

\[\frac{AX}{XB}\cdot\frac{BZ}{ZC}\cdot\frac{CY}{YA}=1 \]

\(\rm Menelaus\)逆定理

若有三点\(X\)\(Y\)\(Z\)分别在边三角形的三边\(AB\)\(BC\)\(CA\)或边的延长线上,并且满足\(\frac{AX}{XB} \cdot \frac{BZ}{ZC} \cdot \frac{CY}{YA}=1\),那么\(X\)\(Y\)\(Z\)三点共线。

(前提:三个点有偶数个点在三角形边上。)

证明:

假设\(X\)\(Y\)\(Z\)三点不共线,直线\(ZY\)\(AB\)交于点\(P\)

根据\(\rm Menelaus\)定理,

\[\frac{AP}{PB}\cdot\frac{BZ}{ZC}\cdot\frac{CY}{YA}=1 \]

\[\rm{已知}\frac{AX}{XB}\cdot\frac{BZ}{ZC}\cdot\frac{CY}{YA}=1 \]

\[\therefore \frac{AP}{PB}=\frac{AX}{XB} \]

\[\therefore P \rm{与} X \rm{重合,即}X\rm{、}Y\rm{、}Z\rm{三点共线} \]

\(\rm Ceva\)定理

在三角形\(\triangle ABC\)任取一点\(O\),延长\(AO\)\(BO\)\(CO\)分别交对边于\(x\)\(y\)\(z\),则有

\[\frac{AX}{XB} \cdot \frac{BZ}{ZC} \cdot \frac{CY}{YA}=1 \]

证明:

\(\therefore \triangle ADC\)被直线\(BE\)所截,

根据\(\rm Menelaus\)定理,

\[\therefore \frac{CB}{BZ}\cdot\frac{ZO}{OA}\cdot\frac{AY}{YC}=1\tag{1} \]

\(\therefore \triangle ABD\)被直线\(CX\)所截,

\[\therefore \frac{BC}{CZ}\cdot\frac{ZO}{OA}\cdot\frac{AX}{XB}=1\tag{2} \]

\[\frac{(2)}{(1)} \rm{得:}\frac{AX}{XB} \cdot \frac{BZ}{ZC} \cdot \frac{CY}{YA}=1 \]

\(\rm Ceva\)逆定理

若有三点\(X\)\(Y\)\(Z\)分别在边三角形的三边\(AB\)\(BC\)\(CA\)或边的延长线上,并且满足\(\frac{AX}{XB} \cdot \frac{BZ}{ZC} \cdot \frac{CY}{YA}=1\),那么\(CX\)\(BY\)\(AZ\)三线共点。

证明:

延长\(CO\)\(AB\)于点\(P\),则有

\[\frac{AP}{PB} \cdot \frac{BZ}{ZC} \cdot \frac{CY}{YA}=1 \]

\[\rm{已知}\frac{AX}{XB} \cdot \frac{BZ}{ZC} \cdot \frac{CY}{YA}=1 \]

\[\therefore \frac{AP}{PB}=\frac{AX}{XB} \]

\[\therefore P \rm{与} X \rm{重合,即}CX\rm{、}BY\rm{、}AZ\rm{三线共点} \]

posted @ 2020-01-27 09:33  XiaoHuang666  阅读(779)  评论(0编辑  收藏  举报