Triangle Centroid & Medians — Simple Guide + CAT Questions

Centroid = common meeting point of the three medians. Think of it as the **balancing point** of a thin triangular plate. Learn the 2:1 rule, quick consequences, and two exam-style problems.

What are medians? Where is the centroid?

A median joins a vertex to the midpoint of the opposite side. The three medians \(AD, BE, CF\) always meet at one point \(G\), called the centroid.

2:1 divide rule: \(G\) divides each median in the ratio \(2:1\) measured from the vertex: \[ \frac{AG}{GD}=\frac{BG}{GE}=\frac{CG}{GF}=\frac{2}{1}. \]
Always inside: The centroid lies inside every triangle.
Balancing point: If the triangle were a uniform sheet, it would balance perfectly at \(G\).
Area fact: The medians cut the triangle into 6 smaller triangles of equal area.
Coordinate shortcut: If \(A(x_1,y_1),B(x_2,y_2),C(x_3,y_3)\), then \[ G\Big(\tfrac{x_1+x_2+x_3}{3},\ \tfrac{y_1+y_2+y_3}{3}\Big). \]

Show concept figure

Medians AD, BE, CF meet at centroid G — Medians \(AD,BE,CF\) intersect at the centroid \(G\) and each is split in \(2:1\) from the vertex.

Why the 2:1 rule (intuitive)

Picture the triangle made of cardboard. A heavier side pulls more, so the “balance point” must sit closer to the vertex than to the midpoint — exactly in the ratio \(2:1\). Formally, using areas (or vectors), each median splits the triangle into equal areas, and the only common point that keeps all three “area balances” consistent is at \(2:1\) from every vertex.

Important clarification (don’t fall for this trap)

If \(AG=BG\), then \(G\) lies on the **perpendicular bisector** of \(AB\). With \(G\) also being the centroid, this forces the triangle to be **isosceles** with \(CA=CB\). But: \(AG=BG\) does not fix the angle \(\angle BGC\) to a constant like \(90^\circ\); it actually depends on the specific triangle.

CAT-style Question 1 — Coordinates drill

Question. Let \(A(0,0)\), \(B(6,0)\), \(C(0,9)\). Find:

Centroid \(G\).
The ratio \(AG:GD\) where \(D\) is midpoint of \(BC\).
Areas of the 6 small triangles formed by the medians.

Solution

1) Centroid: \[ G\Big(\tfrac{0+6+0}{3},\ \tfrac{0+0+9}{3}\Big)=(2,3). \]

2) The 2:1 rule: By definition of centroid, \(AG:GD=2:1\) on median \(AD\).

3) Equal areas: The area of \(\\triangle ABC=\tfrac12\cdot 6\cdot 9=27\). The medians create six equal-area small triangles, each of area \(27/6=\boxed{4.5}\).

CAT-style Question 2 — Using the median length

Question. In \(\\triangle ABC\), let \(a=BC\), \(b=CA\), \(c=AB\). The median from \(A\) to \(BC\) has length \[ m_a=\tfrac12\sqrt{2b^2+2c^2-a^2}. \] If \(AB=AC=10\) and \(BC=12\), find the distance \(AG\) from \(A\) to the centroid \(G\).

Solution

Since \(AB=AC\), from \(A\) the line to midpoint of \(BC\) is a median of length \[ m_a=\tfrac12\sqrt{2\cdot 10^2+2\cdot 10^2-12^2} =\tfrac12\sqrt{200+200-144} =\tfrac12\sqrt{256} =8. \] The centroid divides a median in \(2:1\) from the vertex, so \[ AG=\tfrac{2}{3}\,m_a=\boxed{\tfrac{16}{3}}. \]

Click to view an illustrative centroid sketch

Show sketch

Triangle Centroid & Medians — Simple Guide