Difference between revisions of "Median of a triangle"
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In the following figure, <math>AM</math> is a median of triangle <math>ABC</math>. | 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. | 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. | ||
Revision as of 01:01, 13 June 2022
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
.
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
![$\frac 12 \sqrt{2AB^2+2AC^2-BC^2}$](http://latex.artofproblemsolving.com/3/f/5/3f51d165d3d525f5d9ca4dbed6f8ef7783980ceb.png)
This formula is particularly useful when is right, as by the Pythagorean Theorem we find that
.
See Also
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