Compare the sum of the squares of the median of a triangle to the sum of the squares of sides

Question

You have to compare the sum of the squares of the median of a triangle to the sum of the squares of sides?

Answer 1

$$ 3( AB^2 + BC^2 + AC^2 ) = 4 ( AD^2 + BE^2 + CF^2)$$

AB , BC , CD are lengths of sides of triangle and AD , BE , CF are lengths of medians of triangle

HINT :

start with appollonius theorem and add all

Apollonius Theorem

so you will get relation as

3 (sum of square of sides) = 4 ( sum of squares of medians )

Proof :

$$ AB^2 + AC^2 = 2 ( AD^2 + \frac{BC^2}{4} )$$ $$ AB^2 + BC^2 = 2 ( BE^2 + \frac{AC^2}{4} )$$ $$ AC^2 + BC^2 = 2 ( CF^2 + \frac{AB^2}{4} )$$

Add all above and bring sides to LHS and medians to RHS to get

$$ 3( AB^2 + BC^2 + AC^2 ) = 4 ( AD^2 + BE^2 + CF^2)$$

Answer 2

Let $ABC$ be a triangle and let $I,J,K$ be the midpoints of $[BC],[AC],[AB]$ respectively.

$\overrightarrow{AI}^2=\frac{1}{4}\left(\overrightarrow{AB}+\overrightarrow{AC}\right)^2=\frac{1}{4}\left(AB^2+AC^2+2\overrightarrow{AB}\cdot\overrightarrow{AC}\right)$

The same for $BJ^2$ and $CK^2$. Adding the three equations, we obtain

$$AI^2+BJ^2+CK^2=\frac{1}{4}\left(2AB^2+2AC^2+2BC^2+(\overrightarrow{AB}\cdot\overrightarrow{AC}+\overrightarrow{BA}\cdot\overrightarrow{BC})+(\overrightarrow{AB}\cdot\overrightarrow{AC}+\overrightarrow{CA}\cdot\overrightarrow{CB})+(\overrightarrow{BA}\cdot\overrightarrow{BC}+\overrightarrow{CA}\cdot\overrightarrow{CB})\right)=\frac{3}{4}\left(AB^2+AC^2+BC^2\right)$$

Answer 3

The solution below is probably basically the same as the one by metacompactness. Let our triangle be $ABC$, with the usual conventions about the naming of sidelengths.

Draw the median from vertex $A$ to side $BC$, meeting $BC$ at $M$. By the Cosine Law, we have $$c^2=m^2+\frac{a^2}{4}-2m\cdot\frac{a}{2}\cos(\angle AMB).$$ In the same way, we get $$b^2=m^2+\frac{a^2}{4}-2m\cdot\frac{a}{2}\cos(\angle AMC).$$ Add. The two angles are supplementary, so their cosines have sum $0$. We obtain $$m^2=\frac{2b^2+2c^2-a^2}{4}.$$ We can write down analogous expressions for the squares of the other medians. Add up. The sum of the squares of the medians is $\frac{3}{4}$ times the sum of the squares of the sides.