PlanetMath (more info)
 Math for the people, by the people.
Encyclopedia | Requests | Forums | Docs | Wiki | Random | RSS  
Login
create new user
name:
pass:
forget your password?
Main Menu
Owner confidence rating: Very high Entry average rating: Very high
[parent] common point of triangle medians (Theorem)

Theorem. The three medians of a triangle intersect one another in one point, which divides each median in the ratio $ 2\!:\!1$.


\begin{pspicture}(-1,-0.5)(5,5.5) \psline(4,0)(3,5) \psdot[linecolor=blue](0,0) ... ...r=red](3,5)(2.33,1.67) \psline[linestyle=dotted](2.33,1.67)(2,0) \end{pspicture}
Proof. Let the medians of a triangle $ ABC$ be $ AD$, $ BE$ and $ CF$. Any median vector is the arithmetic mean of the side vectors emanating from the same vertex. Using vectors, let us form three ways all beginning from the vertex $ A$, the first going simply $ 2/3$ of the median vector $ \overrightarrow{AD}$ (blue in the picture):
$\displaystyle \frac{2}{3}\overrightarrow{AD} = \frac{2}{3}\cdot\frac{1}{2}(\ove... ...AB}+\overrightarrow{AC}) = \frac{1}{3}(\overrightarrow{AB}+\overrightarrow{AC})$ (1)

The second way goes first the side vector $ \overrightarrow{AB}$ and then $ 2/3$ of the median vector $ \overrightarrow{BE}$ (green in the picture):
$\displaystyle \overrightarrow{AB}+\frac{2}{3}\overrightarrow{BE} = \overrightar... ...errightarrow{AB})\right] = \frac{1}{3}(\overrightarrow{AB}+\overrightarrow{AC})$ (2)

Similarly, the third way goes first the side vector $ \overrightarrow{AC}$ and then $ 2/3$ of the median vector $ \overrightarrow{CF}$ (red in the picture):
$\displaystyle \overrightarrow{AC}+\frac{2}{3}\overrightarrow{CF} = \overrightar... ...errightarrow{AC})\right] = \frac{1}{3}(\overrightarrow{AB}+\overrightarrow{AC})$ (3)

Thus the ways (2) and (3), where one goes from $ A$ to another vertex and continues along the corresponding median $ 2/3$ of its length, lead to the point $ M$ which is attained directly along $ AD$. This means that all medians intersect in $ M$. The distance of $ M$ from any vertex is $ 2/3$ of the corresponding median, and so the rest of the median is $ 1/3$ of its length, i.e. the ratio of the parts of any median is $ 2\!:\!1$.



"common point of triangle medians" is owned by pahio.
(view preamble)

View style:

See Also: mutual positions of vectors, parallelogram principle, difference of vectors, triangle mid-segment theorem


This object's parent.
Log in to rate this entry.
(view current ratings)

Cross-references: distance, vectors, vertex, side vectors, arithmetic mean, median vector, ratio, divides, point, intersect, triangle
There is 1 reference to this entry.

This is version 5 of common point of triangle medians, born on 2008-02-06, modified 2008-02-08.
Object id is 10241, canonical name is CommonPointOfTriangleMedians.
Accessed 478 times total.

Classification:
AMS MSC51M04 (Geometry :: Real and complex geometry :: Elementary problems in Euclidean geometries)

Pending Errata and Addenda
None.
[ View all 3 ]
Discussion
Style: Expand: Order:
forum policy

No messages.

Interact
post | correct | update request | prove | add result | add corollary | add example | add (any)