Difference between revisions of "Median of a triangle"

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 $\triangle ABC.$

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Difference between revisions of "Median of a triangle"

Latest revision as of 19:24, 6 March 2024

See Also

@@ Line 3: / Line 3: @@
 In the following figure, <math>AM</math> is a median of triangle <math>ABC</math>.
-<center>[[Image:median.PNG]]</center>
+<asy>
-Each triangle has 3 medians.  The medians are [[concurrent]] at the [[centroid]]. The [[centroid]] divides the medians (segments) in a 2:1 ratio.
+import markers;
+pair A, B, C, M;
-[[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>.
+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 ==