Difference between revisions of "Median of a triangle"

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A median of a triangle means either the segment joining one vertex to the midpoint of the opposite side of the triangle, or the straight line that contains this segment. It is a particular case of a [[cevian]] of the triangle.
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A '''median''' of a [[triangle]] is a [[cevian]] of the triangle that joins one [[vertex]] to the [[midpoint]] of the opposite side.
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In the following figure, <math>AM</math> is a median of triangle <math>ABC</math>.
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<asy>
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import markers;
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pair A, B, C, M;
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A = (1, 2);
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B = (0, 0);
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C = (3, 0);
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M = (midpoint(B--C));
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draw(A--B--C--cycle);
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draw(A--M);
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draw(B--M, StickIntervalMarker(1));
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draw(C--M, StickIntervalMarker(1));
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label("$A$", A, N);
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label("$B$", B, W);
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label("$C$", C, E);
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label("$M$", M, S);
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</asy>
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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.
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[[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>
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== See Also ==
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* [[Altitude]]
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* [[Angle bisector]]
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* [[Mass points]]
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* [[Perpendicular bisector]]
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{{stub}}
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[[Category:Definition]]

Latest revision as of 20:24, 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

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