Definition: In geometry, the median of a triangle is a line segment joining a vertex and the midpoint of the opposite side.

Regulations:

1. Every triangle has exactly three medians.

Interactivity

  • Click and drag any vertex of the triangle.
  • Click the checkboxes to show/hide the different medians.
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2. Every median divides the triangle into two smaller triangles of equal area.

Interactivity - A_{\triangle \text{ABAm}} = A_{\triangle \text{ACAm}}%%

  • Click and drag any vertex of the triangle.
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Proof

Open the interactivity above. We want to show: A_{\triangle \text{ABAm}} = A_{\triangle \text{ACAm}}%% .

  • \triangle \text{ABAm} и \triangle \text{ACAm} have the same height h .
  • The area of the triangle \triangle \text{ABAm} is:   A_{\triangle \text{ABAm}} = \frac{1}{2} h \cdot \overline{BAm} .
  • The area of the triangle \triangle \text{ACAm} is:   A_{\triangle \text{ACAm}} = \frac{1}{2}\, h \cdot \overline{CAm} .
  • Because \overline{АAm} is a median, the bases are the same, i.e. \overline{BAm}=\overline{BAm} .
    • It follows: \frac{1}{2} h \cdot \overline{BAm} = \frac{1}{2}\, h \cdot \overline{CAm} , that is
    • A_{\triangle \text{ABAm}} = A_{\triangle \text{ACAm}}%% , as we wanted.

3. The three medians intersect in a common point T called the centroid of the triangle.

Interactivity

  • Click and drag any vertex of the triangle.
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Discussion about proof

  • The proof is a direct application of Ceva's theorem and the definition of median.
  • From this it follows that the centroid is always an internal point of the triangle.

4. The three medians divide the triangle into 6 smaller triangles of equal area.

Interactivity

  • Click and drag any vertex of the triangle. Click the checkboxes to show/hide the various triangles.
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Proof

1. Using the same proof as in regulation 2 (above), the areas of T1 and T2 are equal, the areas of T3 and T4 are equal and the areas of T5 and T6 are equal.

That is А_{T1}=А_{T2} and А_{T3}=А_{T4} and А_{T5}=А_{T6}

2. It remains to show that А_{T1} = А_{T3} = А_{T5}

Because the proofs are analogous (the same with different numbers), it is enough to prove any one equality.
We will show А_{T1} = А_{T5}

3. According to regulation 2, the areas of the triangles ABAm and ACAm are equal.

So, А_{T1}+А_{T2}+А_{T3} = А_{T4}+А_{T5}+А_{T6}

4. Substitution from (2) into (1), we have А_{T1}+А_{T1}+А_{T3} = А_{T3}+А_{T5}+А_{T5} so that

2 \cdot А_{T1} = 2 \cdot А_{T5} or
А_{T1} = А_{T5} , which is what we wanted (2)

Conclusion All six little triangles have the same area.

 

5. Two-thirds of the length of the median is between the vertex and the centroid; one-third is between the centroid and the midpoint.

Interactivity

  • Click and drag any vertex of the triangle. Notice how the part of the median towards the vertex is always 2 times longer than the part towards the midpoint. (If it is not "exactly" twice, increase the number of decimals through the command Options -> Decimal places.)
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Page last modified on September 14, 2008, at 11:06 PM