Six Equal-Area Triangles

An animated proof that the three medians of a triangle divide it into six smaller triangles of equal area.

Keywords: centroid, medians, equal area, triangle partition, six triangles, 2:1 ratio

Prerequisites: centroid-proof, area-proof · Difficulty: intermediate

The three medians of a triangle divide it into six smaller triangles, and remarkably all six have the same area — each is exactly $\frac{1}{6}$ of the original triangle.

Strategy: a median halves the area (equal bases, shared height), and the centroid $G$ splits each median in ratio $2 : 1$ from the vertex. The animation combines these to show each sub-triangle is one-third of a half, hence $\frac{1}{6} [A B C]$ , then extends the result to all six by symmetry.

Step 1: A Median Halves the Area

A median connects a vertex to the midpoint of the opposite side. It divides the triangle into two smaller triangles with equal area.

Why? The two sub-triangles share the same height (from the vertex), and their bases are equal (since the median lands at the midpoint).

[△ A B M_{A}] = [△ A C M_{A}] = \frac{1}{2} [△ A B C]

Step 2: The Centroid's $2 : 1$ Ratio

Draw the other two medians. All three meet at a single point — the centroid $G$ .

By the centroid theorem (proven in the previous section), $G$ divides each median in ratio $2 : 1$ from the vertex:

\frac{A G}{G M _{A}} = \frac{B G}{G M _{B}} = \frac{C G}{G M _{C}} = \frac{2}{1}

We'll combine this with the "median halves area" fact to compute the area of each sub-triangle directly.

Step 3: Six Sub-Triangles of Equal Area

The three medians partition $△ A B C$ into six smaller triangles meeting at $G$ . Despite their different shapes, all six have exactly the same area:

[△_{1}] = [△_{2}] = \dots = [△_{6}] = \frac{1}{6} [△ A B C]

Step 4: One Sub-Triangle Has Area $\frac{1}{6} [A B C]$

We compute the area of one sub-triangle, $△ A G M_{C}$ , directly from the two facts above.

Median $C M_{C}$ cuts $△ A B C$ into two halves; the half containing $A$ is $△ A C M_{C}$ , with area $\frac{1}{2} [A B C]$ .
Inside that half, the segment $A G$ splits it into $△ A C G$ and $△ A G M_{C}$ . These share vertex $A$ , and their bases $C G$ and $G M_{C}$ sit on the same line.
Since $C G : G M_{C} = 2 : 1$ (the centroid theorem), the two areas are in the same ratio $2 : 1$ .

Therefore $△ A G M_{C}$ is one-third of the half:

[△ A G M_{C}] = \frac{1}{3} \cdot \frac{1}{2} [A B C] = \frac{1}{6} [A B C]

Step 5: All Six Sub-Triangles Are Equal

The same argument applies to every sub-triangle. For each one, we pick the median whose half contains it, and the centroid's $2 : 1$ split on that median gives the same answer:

[△ A G M_{C}] = [△ G M_{C} B] = [△ B G M_{A}] = [△ G M_{A} C] = [△ C G M_{B}] = [△ G M_{B} A] = \frac{1}{6} [A B C]

All six sub-triangles have the same area — another manifestation of the centroid as the triangle's center of mass.

The three medians divide triangle $A B C$ into six sub-triangles, each with area exactly $\frac{1}{6}$ of the original:

[A G M_{C}] = [G M_{C} B] = [B G M_{A}] = [G M_{A} C] = [C G M_{B}] = [G M_{B} A]

= \frac{1}{6} [A B C]

This elegant partition is another manifestation of the centroid as the triangle's center of mass. ∎

Notes

Notation. Throughout this proof we write $[X Y Z]$ for the area of polygon $X Y Z$ .