Difference between revisions of "Median of a triangle"

Revision as of 14:58, 4 April 2020

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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-A median of a [[triangle]] is a [[cevian]] of a triangle that goes from  either the segment joining one vertex to the midpoint of the opposite side of the triangle, or the straight line that contains this segment.
+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, <math>AM</math> is a median of triangle <math>ABC</math>.
+<center>[[Image:median.PNG]]</center>
+Each triangle has <math>3</math> medians.  The medians are [[concurrent]] at the [[centroid]]. The [[centroid]] divides the medians (segments) in a <math>2:1</math> 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>.
-The medians are [[concurrent]] at the [[centroid]]. The [[centroid]] divides the medians in a 2:1 ratio.
 == See Also ==
-*[[Cevian]]
+* [[Altitude]]
-*[[Angle bisector]]
+* [[Angle bisector]]
-*[[Perpendicular bisector]]
+* [[Mass points]]
-*[[Altitude]]
+* [[Perpendicular bisector]]
+{{stub}}
+[[Category:Definition]]