Art of Problem Solving
Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Median of a triangle: Difference between revisions

← Older edit
MCrawford (talk | contribs)
1,942 edits
wikified and added mass points to "See Also"
Clever14710owl (talk | contribs)
editor
38 edits
No edit summary
 
(7 intermediate revisions by 5 users not shown)
Line 1: Line 1:
A '''median''' of a [[triangle]] is a [[cevian]] of the 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.  


The medians are [[concurrent]] at the [[centroid]]. The [[centroid]] divides the medians (segments) in a 2:1 ratio.
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 ==  
Line 9: Line 32:
* [[Mass points]]
* [[Mass points]]
* [[Perpendicular bisector]]
* [[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, $AM$ is a median of triangle $ABC$.

[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 $3$ 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 $BM=AM=CM$. This occurs when $M$ is the circumcenter of $\triangle ABC.$

See Also

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=Median_of_a_triangle&oldid=216546"