Midsegments and Medians

$\overline{MN} \parallel \overline{BC} \qquad \text{and} \qquad MN = \frac{1}{2} BC$

Proof of the Midsegment TheoremLink to midsegments-proof-of-the-midsegment-theorem

Place the triangle in the coordinate plane with:

$A = (0,\, 2a), \quad B = (-2b,\, 0), \quad C = (2c,\, 0)$

Then the midpoints are:

$M = \frac{A + B}{2} = (-b,\, a), \qquad N = \frac{A + C}{2} = (c,\, a)$

The slope of $\overline{MN}$:

$\text{slope of } MN = \frac{a - a}{c - (-b)} = 0$

The slope of $\overline{BC}$:

$\text{slope of } BC = \frac{0 - 0}{2c - (-2b)} = 0$

Equal slopes confirm $\overline{MN} \parallel \overline{BC}$. For the length:

$MN = c - (-b) = b + c, \qquad BC = 2c - (-2b) = 2(b+c)$

$\boxed{MN = \frac{1}{2} BC}$

The Medial TriangleLink to midsegments-the-medial-triangle

Connecting all three midsegments produces the medial triangle, which partitions $\triangle ABC$ into four congruent triangles, each similar to the original with ratio $1:2$:

$[\triangle_{\text{medial}}] = \frac{1}{4}[\triangle ABC]$

MediansLink to medians

A median connects a vertex to the midpoint of the opposite side. Every triangle has exactly three medians.

Finding the CentroidLink to medians-finding-the-centroid

For a triangle with vertices $A$, $B$, $C$, the centroid is simply their average:

$G = \frac{A + B + C}{3}$

In coordinates with $A = (x_1, y_1)$, $B = (x_2, y_2)$, $C = (x_3, y_3)$:

$G = \left(\frac{x_1 + x_2 + x_3}{3},\; \frac{y_1 + y_2 + y_3}{3}\right)$

The 2

RatioLink to medians-the-2-ratio

The centroid does not sit at the midpoint of each median — it divides each one in the ratio $2:1$ from vertex to midpoint:

$\frac{AG}{GM'} = \frac{2}{1}$

where $M'$ is the midpoint of $\overline{BC}$. More precisely, if $M' = \frac{B+C}{2}$, then:

$G = A + \frac{2}{3}(M' - A) = \frac{A + B + C}{3}$

This can be verified for all three medians simultaneously — the same point $G$ satisfies the $2:1$ condition on each one.

Connecting Midsegments and MediansLink to connecting-midsegments-and-medians

The medians and midsegments are more tightly linked than they first appear. Each median of $\triangle ABC$ is also a median of the medial triangle — and the centroid $G$ is shared by both.

Furthermore, the three midsegments divide $\triangle ABC$ into four triangles. The centroid $G$ of the original triangle coincides with the centroid of the medial triangle, since:

$G = \frac{A + B + C}{3} = \frac{M_{AB} + M_{BC} + M_{CA}}{3} \cdot \frac{?}{?}$

where $M_{AB} = \frac{A+B}{2}$, $M_{BC} = \frac{B+C}{2}$, $M_{CA} = \frac{C+A}{2}$. Summing:

$M_{AB} + M_{BC} + M_{CA} = A + B + C$

$\boxed{\text{centroid of medial triangle} = \frac{A+B+C}{3} = G}$

SummaryLink to summary