Difference between revisions of "Median of a triangle"
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| − | A median of a [[triangle]] is a [[cevian]] of | + | 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>. | ||
| + | |||
| + | <asy> | ||
| + | import markers; | ||
| + | pair A, B, C, M; | ||
| + | |||
| + | A = (1, 2); | ||
| + | B = (0, 0); | ||
| + | C = (3, 0); | ||
| + | M = (midpoint(B--C)); | ||
| + | |||
| + | draw(A--B--C--cycle); | ||
| + | draw(A--M); | ||
| + | draw(B--M, StickIntervalMarker(1)); | ||
| + | draw(C--M, StickIntervalMarker(1)); | ||
| + | |||
| + | label("$A$", A, N); | ||
| + | label("$B$", B, W); | ||
| + | label("$C$", C, E); | ||
| + | label("$M$", M, S); | ||
| + | </asy> | ||
| + | |||
| + | 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>. This occurs when <math>M</math> is the circumcenter of <math>\triangle ABC.</math> | ||
| − | |||
== See Also == | == See Also == | ||
| − | *[[ | + | * [[Altitude]] |
| − | *[[Angle bisector]] | + | * [[Angle bisector]] |
| − | *[[Perpendicular bisector]] | + | * [[Mass points]] |
| − | + | * [[Perpendicular bisector]] | |
| + | {{stub}} | ||
| + | [[Category:Definition]] | ||
Latest revision as of 19:24, 6 March 2024
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, 
is a median of triangle 
.
![[asy] import markers; pair A, B, C, M; A = (1, 2); B = (0, 0); C = (3, 0); M = (midpoint(B--C)); draw(A--B--C--cycle); draw(A--M); draw(B--M, StickIntervalMarker(1)); draw(C--M, StickIntervalMarker(1)); label("$A$", A, N); label("$B$", B, W); label("$C$", C, E); label("$M$", M, S); [/asy]](http://latex.artofproblemsolving.com/a/e/5/ae531958c09b155c0dca21e9cd9b9bb01d391fb9.png)
Each triangle has 
medians. The medians are concurrent at the centroid. The centroid divides the medians (segments) in a 
ratio.
Stewart's Theorem applied to the case 
, gives the length of the median to side 
equal to
This formula is particularly useful when 
is right, as by the Pythagorean Theorem we find that 
. This occurs when 
is the circumcenter of 
See Also
This article is a stub. Help us out by expanding it.
