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. Remarkably, all six have the same area — each is exactly $\frac{1}{6}$ of the original triangle.

Notation. Throughout this proof we write $[XYZ]$ for the area of polygon $XYZ$.

Proof Strategy

The proof combines two facts:

With these in hand:

  1. Pick any sub-triangle, say $\triangle AGM_C$.
  2. It sits inside the half $\triangle ACM_C$ (cut off by median $CM_C$), whose area is $\tfrac{1}{2}[ABC]$.
  3. The segment $AG$ splits this half into $\triangle ACG$ and $\triangle AGM_C$. They share vertex $A$, and their bases $CG$ and $GM_C$ sit on the same line with $CG : GM_C = 2 : 1$.
  4. So $[\triangle AGM_C] = \tfrac{1}{3} \cdot \tfrac{1}{2}[ABC] = \tfrac{1}{6}[ABC]$.
  5. By the same argument at each vertex, all six sub-triangles have area $\tfrac{1}{6}[ABC]$.

The Proof

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).

[\triangle ABM_A] = [\triangle ACM_A] = \frac{1}{2}[\triangle ABC]

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{AG}{GM_A} = \frac{BG}{GM_B} = \frac{CG}{GM_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 $\triangle ABC$ into six smaller triangles meeting at $G$. Despite their different shapes, all six have exactly the same area:

[\triangle_1] = [\triangle_2] = \cdots = [\triangle_6]
    = \tfrac{1}{6}[\triangle ABC]

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

We compute the area of one sub-triangle, $\triangle AGM_C$, directly from the two facts above.

  1. Median $CM_C$ cuts $\triangle ABC$ into two halves; the half containing $A$ is $\triangle ACM_C$, with area $\tfrac{1}{2}[ABC]$.
  2. Inside that half, the segment $AG$ splits it into $\triangle ACG$ and $\triangle AGM_C$. These share vertex $A$, and their bases $CG$ and $GM_C$ sit on the same line.
  3. Since $CG : GM_C = 2 : 1$ (the centroid theorem), the two areas are in the same ratio $2:1$.

Therefore $\triangle AGM_C$ is one-third of the half:

[\triangle AGM_C]
      = \tfrac{1}{3} \cdot \tfrac{1}{2}[ABC]
      = \tfrac{1}{6}[ABC]

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:

[\triangle AGM_C] = [\triangle GM_CB] = [\triangle BGM_A]
      = [\triangle GM_AC] = [\triangle CGM_B] = [\triangle GM_BA]
      = \tfrac{1}{6}[ABC]

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

Conclusion

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

[AGM_C] = [GM_CB] = [BGM_A] = [GM_AC] = [CGM_B] = [GM_BA]
    
= \frac{1}{6} [ABC]

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