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Mathematics LibreTexts

8.3: Medians and centroid

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    23628
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    A median of a triangle is the segment joining a vertex to the midpoint of the opposing side.

    Theorem \(\PageIndex{1}\)

    The three medians of any nondegenerate triangle intersect in a single point. Moreover, the point of intersection divides each median in the ratio 2:1.

    The point of intersection of medians is called the centroid of the tri- angle; it is usually denoted by M. In the proof we will apply Exercise 3.4.3 and Exercise7.3.1; their complete solutions are given in the hits.

    Proof

    Consider a nondegenerate triangle \(ABC\). Let \([AA']\) and \([BB']\) be its medians. According to Exercise 3.4.3, \([AA']\) and \([BB']\) have a point of intersection; denote it by \(M\).

    截屏2021-02-15 下午3.30.48.png

    Draw a line \(\ell\) thru \(A'\) parallel to \((BB')\). Applying Exercise7.3.1 for \(\triangle BB'C\) and \(\ell\), we get that \(\ell\) cross \([B'C]\) at some point \(X\) and

    \(\dfrac{CX}{CB'} = \dfrac{CA'}{CB} = \dfrac{1}{2};\)

    that is, \(X\) is the midpoint of \([CB']\).

    Since \(B'\) is the midpoint of \([AC]\) and \(X\) is the midpoint of \([B'C]\), we get that

    \(\dfrac{AB'}{AX} = \dfrac{2}{3}.\)

    Applying Exercise 7.3.1 for \(\triangle XA'A\) and the line \((BB')\), we get that

    \[\dfrac{AM}{AA'} = \dfrac{AB'}{AX} = \dfrac{2}{3};\]

    that is, \(M\) divides \([AA']\) in the ratio 2:1.

    Note that 8.3.1 uniquely defines \(M\) on \([AA']\). Repeating the same argument for medians \([AA']\) and \([CC']\), we get that they intersect at \(M\) as well, hence the result.

    Exercise \(\PageIndex{1}\)

    Let \(\square ABCD\) be a nondegenerate quadrangle and \(X, Y, V, W\) be the midpoints of its sides \([AB], [BC], [CD]\), and \([DA]\). Show that \(\square XYVW\) is a parallelogram.

    Hint

    Use the idea from the proof of Theorem \(\PageIndex{1}\) to show that \((XY) \parallel (AC) \parallel (VW)\) and \((XV) \parallel (BD) \parallel (YW)\).