Triangle Medians, Midpoints, and the Centroid
In any triangle, a median is a segment drawn from a vertex to the midpoint of the
opposite side. If the triangle has vertices \(A,B,C\), then the median from \(A\) meets side \(BC\) at the point
\(M_a\), where \(M_a\) is the midpoint of \(BC\). Similarly, the medians from \(B\) and \(C\) meet sides \(AC\) and
\(AB\) at midpoints \(M_b\) and \(M_c\). Medians are important because they describe a triangle’s “balance structure”:
if you cut a uniform triangular plate, its balancing point is the centroid, and it lies on each median.
A key fact is that the three medians are concurrent, meaning they intersect at exactly one point.
That intersection point is called the centroid, usually denoted \(G\). The centroid has a famous ratio
property: it divides every median in the ratio \(2:1\), measured from the vertex. In other words, on median
\(\overline{AM_a}\), the centroid satisfies \(AG = \tfrac{2}{3}AM_a\) and \(GM_a = \tfrac{1}{3}AM_a\). This explains why
the centroid is sometimes described as the triangle’s “center of mass” for a uniform density triangle.
When only side lengths are known, medians can still be computed directly using a classical formula (often derived from
the Law of Cosines or from Apollonius’ Theorem). If side \(a\) is opposite vertex \(A\) (so \(a = |BC|\)), and
\(b = |AC|\), \(c = |AB|\), then the median length from \(A\) is
\[
m_a = \frac12\sqrt{2b^2 + 2c^2 - a^2}.
\]
By symmetry,
\[
m_b = \frac12\sqrt{2a^2 + 2c^2 - b^2},\qquad
m_c = \frac12\sqrt{2a^2 + 2b^2 - c^2}.
\]
These expressions are useful because they require only the side lengths and automatically incorporate triangle shape.
For visualization, a triangle can be placed on a coordinate plane without changing its geometry. A convenient choice is
to set \(A=(0,0)\) and \(B=(c,0)\), so segment \(\overline{AB}\) lies on the \(x\)-axis with length \(c\). The third
vertex \(C\) can then be determined from the distances \(|AC|=b\) and \(|BC|=a\). Solving those distance conditions gives
\[
x_C = \frac{b^2 + c^2 - a^2}{2c},\qquad
y_C = \sqrt{b^2 - x_C^2}.
\]
Once the vertices are known, midpoints follow from averaging endpoints, and the centroid is the average of the three
vertices:
\[
G=\left(\frac{x_A+x_B+x_C}{3},\,\frac{y_A+y_B+y_C}{3}\right).
\]
Finally, always check the triangle inequality: each side must be shorter than the sum of the other two.
If \(a+b\le c\) (or any similar inequality fails), a triangle cannot be formed, so median
