Menelaus$524948$ - traduzione in greco
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Menelaus$524948$ - traduzione in greco

THEOREM
Theorem of Menelaus; Menelaus theorem; Menelaos's theorem; Menelaus' theorem
  • Menelaus's theorem, case 2: line DEF is entirely outside triangle ABC
  • Menelaus's theorem, case 1: line DEF passes inside triangle ABC

Menelaus      
n. μενέλαος

Wikipedia

Menelaus's theorem

Menelaus's theorem, named for Menelaus of Alexandria, is a proposition about triangles in plane geometry. Suppose we have a triangle ABC, and a transversal line that crosses BC, AC, and AB at points D, E, and F respectively, with D, E, and F distinct from A, B, and C. A weak version of the theorem states that

| A F | | F B | × | B D | | D C | × | C E | | E A | = 1 , {\displaystyle {\frac {|AF|}{|FB|}}\times {\frac {|BD|}{|DC|}}\times {\frac {|CE|}{|EA|}}=1,}

where |AB| is taken to be the ordinary length of segment AB: a positive value.

The theorem can be strengthened to a statement about signed lengths of segments, which provides some additional information about the relative order of collinear points. Here, the length AB is taken to be positive or negative according to whether A is to the left or right of B in some fixed orientation of the line; for example, AF/FB is defined as having positive value when F is between A and B and negative otherwise. The signed version of Menelaus's theorem states

A F F B × B D D C × C E E A = 1. {\displaystyle {\frac {AF}{FB}}\times {\frac {BD}{DC}}\times {\frac {CE}{EA}}=-1.}

Equivalently,

A F × B D × C E = F B × D C × E A . {\displaystyle AF\times BD\times CE=-FB\times DC\times EA.}

Some authors organize the factors differently and obtain the seemingly different relation

F A F B × D B D C × E C E A = 1 , {\displaystyle {\frac {FA}{FB}}\times {\frac {DB}{DC}}\times {\frac {EC}{EA}}=1,}

but as each of these factors is the negative of the corresponding factor above, the relation is seen to be the same.

The converse is also true: If points D, E, and F are chosen on BC, AC, and AB respectively so that

A F F B × B D D C × C E E A = 1 , {\displaystyle {\frac {AF}{FB}}\times {\frac {BD}{DC}}\times {\frac {CE}{EA}}=-1,}

then D, E, and F are collinear. The converse is often included as part of the theorem. (Note that the converse of the weaker, unsigned statement is not necessarily true.)

The theorem is very similar to Ceva's theorem in that their equations differ only in sign. By re-writing each in terms of cross-ratios, the two theorems may be seen as projective duals.