How can we find the length of the median, m, given the length of the three sides of a \triangle ABC a,b and c?
HINT for STEP 1 - Can you turn \triangle ABC into a parallelogram ABA'C? Try this first, then read on to see how you went.
By using \triangle ABC and recognising that this forms exactly half of the parallelogram, we copy the triangle and rotate it 180 degrees.
This results in the following parallelogram.
From here can you identify all the equalities and similarities that exist? There are a lot! Being able to identify these really helps in trying to construct proofs or logical arguments. Particularly can you see the lengths of the 2 diagonals?
We can see that the diagonal BC has length a, as we were told the lengths in the original statement. Also we can see the diagonal AA' has length 2m.
There is a property of parallelograms that states that the sum of squares of the sides of a parallelogram is equal to the sum of the squares of the diagonals, (if you wish to see this proof go here).
So we can say then that (2m)^2+ a^2 = 2b^2 + 2c^2
This means that the length of the median, m is
m_a =\frac{ \sqrt{2b^2 + 2c^2 - a^2}}{2}
We can follow the same process and find the medians m_b, and m_c. Can you find these expressions? Try this before continuing on to check your answer.
m_b =\frac{ \sqrt{2c^2 + 2a^2 - b^2}}{2}
m_c =\frac{ \sqrt{2a^2 + 2b^2 - c^2}}{2}
What is the sum of the squares of the medians?
| m_a^2 + m_b^2 + m_c^2 | = | (\frac{ \sqrt{2b^2 + 2c^2 - a^2}}{2} )^2+ (\frac{ \sqrt{2c^2 + 2a^2 - b^2}}{2})^2 + (\frac{ \sqrt{2a^2 + 2b^2 - c^2}}{2})^2 |
| = | (\frac{ 2b^2 + 2c^2 - a^2}{4} )+ (\frac{ 2c^2 + 2a^2 - b^2}{4}) + (\frac{2a^2 + 2b^2 - c^2}{4}) | |
| = | \frac{3a^2+3b^2+3c^2}{4} | |
| = | \frac{3}{4}\left(a^2+b^2+c^2\right) |
Whilst this doesn't really have any real life applications, it is a rather neat result. That for any triangle, the sum of the squares of the medians is 3 quarters of the sum of the squares of the sides.