Difference between revisions of "Median of a triangle"

m (bah, I hate stubs)
(Added median-length formula.)
Line 5: Line 5:
 
<center>[[Image:median.PNG]]</center>
 
<center>[[Image:median.PNG]]</center>
 
Each triangle has 3 medians.  The medians are [[concurrent]] at the [[centroid]]. The [[centroid]] divides the medians (segments) in a 2:1 ratio.
 
Each triangle has 3 medians.  The medians are [[concurrent]] at the [[centroid]]. The [[centroid]] divides the medians (segments) in a 2:1 ratio.
 +
 +
[[Stewart's theorem]] applied to the case <math>m=n</math>, gives the length of the median to side <math>BC</math> equal to <center><math>\frac 12 \sqrt{2AB^2+2AC^2-BC^2}</math></center> This formula is particularly useful when <math>\angle CAB</math> is right, as by the Pythagorean Theorem we find that <math>BM=AM=CM</math>.
  
 
== See Also ==  
 
== See Also ==  

Revision as of 08:15, 21 November 2006

A median of a triangle is a cevian of the triangle that joins one vertex to the midpoint of the opposite side.

In the following figure, $AM$ is a median of triangle $ABC$.

Median.PNG

Each triangle has 3 medians. The medians are concurrent at the centroid. The centroid divides the medians (segments) in a 2:1 ratio.

Stewart's theorem applied to the case $m=n$, gives the length of the median to side $BC$ equal to

$\frac 12 \sqrt{2AB^2+2AC^2-BC^2}$

This formula is particularly useful when $\angle CAB$ is right, as by the Pythagorean Theorem we find that $BM=AM=CM$.

See Also

This article is a stub. Help us out by expanding it.