Median of a triangle: Difference between revisions

Revision as of 08:16, 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 $ABC$ .

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 $BM=AM=CM$ .

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 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>.
+[[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 ==