So I stumbled upon a question, if we are given the lengths of all the medians of a triangle, can we find its sides?!
Let ABC be a triangle, with D, E, F midpoints of \(\overlinesegment{BC}, \overlinesegment{CA}, \overlinesegment{AC}\) respectively.
Thus let, \(m_1 = \overlinesegment{AD}, m_2 = \overlinesegment{BE}, m_3 = \overlinesegment{CF}\) denote the length of the respective medians.
and let \(BD = DC = x\), \(CE = EA = y\), \(AF = FB = z\),

By Apollonius's Theorem or Stewart's Theorem, we have
\(4y^2 + 4z^2 = 2x^2 + 2m_1^2\)
\(\implies 2y^2 + 2z^2 = x^2 + m_1^2\)

By symmetry we have:
\(2z^2 + 2x^2 = y^2 + m_2^2\)
\(2x^2 + 2y^2 = z^2 + m_3^2\)

Adding all the equations, we get:
\(3(x^2 + y^2 + z^2) = m_1^2 + m_2^2 + m_3^2 \)
This is an important result, which I will talk about later!

Substituting the above result into any of the 3 equations, we can find the sides!
We have:
\((3x)^2 = 2m_2^2 + 2m_3^2 - m_1^2\)
\((3y)^2 = 2m_3^2 + 2m_1^2 - m_2^2\)
\((3z)^2 = 2m_1^2 + 2m_2^2 - m_3^2\)

So now we have:
\(BD^2 = CD^2 = \frac{2m_2^2 + 2m_3^2 - m_1^2}{9}\)

And so we have successfully found the sides of the triangle, and in the process we have found 3 important results.

• 3 times the sum of squares of the side lengths equal 4 times the sum of squares of medians.
• The square of half the opposite side from any vertex equals twice the sum of squares of medians from the other vertices subtracted by the square of median from the vertex with everything divided by 9
• Since we have:
  \(BD^2 = CD^2 = \frac{2m_2^2 + 2m_3^2 - m_1^2}{9}\)
  \(\implies\) since \(BD^2\) and \(CD^2\) are non zero real numbers
  \(\implies 2m_2^2 + 2m_3^2 \gt m_1^2\)
  Thus twice the sum of squares of any two medians of a triangle is greater than the square of the third median.