Related Directory Categories
- Geometry (509)
- Algebraic Geometry (121)
- Sacred Geometry and Art (68)
- Computational Geometry (68)
- Events (64)
- Polytopes (58)
- Geometri (55)
- Designers and Galleries (54)
- Past Events (53)
- Events (52)
Sources
Help build the largest human-edited directory on the web.
Submit a site - Become an editor
Wikipedia. Licensed under the GNU Free Documentation License.
Are you an expert in this subject? Join the discussion and share your knowledge at Wikipedia.org.
Median (geometry)
From Wikipedia, the free encyclopedia
In geometry, a median of a triangle is a cevian joining a vertex to the midpoint of the opposing side.
Every triangle has exactly three medians; one running from each vertex to the opposite side.
Contents |
The three medians are concurrent at a point known as the triangle's centroid, which or center of mass of the triangle. Note that this means that the centroid is always in the interior of the triangle. Two-thirds of the length of each median is between the vertex and the centroid, while one-third is between the centroid and the midpoint of the opposite side.
It can be proven that the three medians are concurrent using Ceva's theorem. Since the medians of a triangle pass through the midpoint of the sides, it's clear that ratio between the two segments they are divided into have ratios of 1. Thus, since 
, the three medians are concurrent.
The latter point mentioned can be proven using the midsegment theorem.
The three medians divide the triangle into six smaller triangles of equal area.
Any other lines which divide the area of the triangle into two equal parts do not pass through the centroid.
Consider a triangle ABC. Let D be the midpoint of 
, E be the midpoint of 
, Let D be the midpoint of , F be the midpoint of 
, and O be the centroid.
By definition, AD = DB,AF = FC,BE = EC, thus [ADO] = [BDO],[AFO] = [CFO],[BEO] = [CEO],[ABE] = [ACE], where [ABC] represents the area of triangle 
.
We have:
- [ABO] = [ABE] − [BEO]
- [ACO] = [ACE] − [CEO]
Thus, [ABO] = [ACO] and ![[ADO]=[DBO], [ADO]=\frac{1}{2}[ABO]](http://upload.wikimedia.org/math/5/9/8/598b6f25309c905043792c94accee187.png)
Since ![[AFO]=[FCO], [AFO]= \frac{1}{2}AFO=\frac{1}{2}[ABO]=[ADO]](http://upload.wikimedia.org/math/9/0/c/90ce9a758865acd5b552ca49ed358613.png)
, therefore, [AFO] = FCO = [ABO] = [ADO]. Using the same method, you can show that [AFP] = [FCO] = [ABO] = [ADO] = [BEO] = [CEO].
Applying Stewart's theorem one gets:
where a is the side of the triangle whose midpoint is the extreme point of median m.
- Medians and Area Bisectors of a Triangle
- The Medians at cut-the-knot
- Area of Median Triangle at cut-the-knot
- Medians of a triangle With interactive animation
- Constructing a median of a triangle with compass and straightedge animated demonstration