Median of a triangle: Difference between revisions
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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 00:01, 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
This formula is particularly useful when
is right, as by the Pythagorean Theorem we find that
.
See Also
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