Menelaus theorem

case 1

Case 2

The Menelaus' theorem , named after the Greek mathematician Menelaus ( Alexandria , about 100 n. Chr.), Makes a statement about straight lines , the triangles overlap.

Let a triangle ABC and a straight line be given which intersects the triangle sides [BC], [CA] and [AB] or their extensions at points X, Y and Z. Then:

${\displaystyle TV(B,C,X)\cdot TV(C,A,Y)\cdot TV(A,B,Z)=-1}$

Conversely, one can conclude from the correctness of this relationship that the points X, Y and Z lie on a straight line.

Here, the division ratio of which three lie on a straight points to is defined by . If is between and , this part ratio is the same , otherwise the same . ${\displaystyle TV(U,V,W)}$ ${\displaystyle U,V,W}$ ${\displaystyle U,V,W}$ ${\displaystyle U\neq V}$ ${\displaystyle W=U+TV(U,V,W)\cdot (V-W)}$ ${\displaystyle W}$ ${\displaystyle U}$ ${\displaystyle V}$ ${\displaystyle {\overline {UW}}/{\overline {WV}}}$ ${\displaystyle -{\overline {UW}}/{\overline {WV}}}$

If you only consider the lengths of the route, you can also write the above equation in the following form:

${\displaystyle {\overline {AZ}}\cdot {\overline {BX}}\cdot {\overline {CY}}={\overline {AY}}\cdot {\overline {BZ}}\cdot {\overline {CX}}}$

Since orientation is lost here, this equation is not sufficient to reverse the theorem, cf. Theorem of Ceva .

proof

To prove the theorem

Menelaus' theorem can be proved with the help of the ray theorem. Consider three perpendiculars on the given straight line, which start from corners A, B and C. The lengths of the plumb lines are denoted by , and . ${\displaystyle a}$ ${\displaystyle b}$ ${\displaystyle c}$

The following equations are obtained from the theorem of rays:

${\displaystyle {\overline {AZ}}:{\overline {BZ}}=a:b}$

${\displaystyle {\overline {BX}}:{\overline {CX}}=b:c}$

${\displaystyle {\overline {CY}}:{\overline {AY}}=c:a}$

If you multiply these three equations together, the result is

${\displaystyle {\frac {\overline {AZ}}{\overline {BZ}}}\cdot {\frac {\overline {BX}}{\overline {CX}}}\cdot {\frac {\overline {CY}}{\overline {AY}}}={\frac {a}{b}}\cdot {\frac {b}{c}}\cdot {\frac {c}{a}}={\frac {a\cdot b\cdot c}{b\cdot c\cdot a}}=1}$

and further (by multiplying by the denominator)

${\displaystyle {\overline {AZ}}\cdot {\overline {BX}}\cdot {\overline {CY}}={\overline {AY}}\cdot {\overline {BZ}}\cdot {\overline {CX}}}$.

application

Menelaus' theorem, together with its converse, provides a criterion for collinear points. One consequence is Ceva's theorem .

literature

• Hans Schupp: Elementary Geometry. Schöningh, Paderborn 1977, ISBN 3-506-99189-2 , p. 124 ff., P. 136 ( Uni-Taschenbücher 669 Mathematik )
• Max Koecher , Aloys Krieg : level geometry . 3. Edition. Springer-Verlag, Berlin 2007, ISBN 978-3-540-49327-3 , pp. 78-81
• Branko Grunbaum, GC Shephard: Ceva, Menelaus, and the Area Principle . In: Mathematics Magazine , Volume 68, No. 4, Oct. 1995, pp. 254-268 ( JSTOR 2690569 )

Web links

Commons : Menelaos's theorem  - collection of images, videos and audio files

• Eric W. Weisstein : Menelaus' Theorem . In: MathWorld (English).
• Menelaus 'theorem an animated, geometric proof of Menelaus' theorem
• Paul Yiu: Euclidean Geometry Notes (PDF) script, Florida Atlantic University, pp. 87-102