Center Centroid

Simple Method: Consider ABC as equilateral triangle of side 1 unit. then the calculation becomes simple.

Let $AB=c$ , $AC=b$ and $BC=a$ . Also, let $E$ , $F$ and $G$ the the intersection points of $AD$ with $BC$ , $BD$ with $AC$ and $CD$ with $AB$ , respectively.

By Apollonius' theorem, we know that:

$AE=\dfrac{\sqrt{2(b^2+c^2)-a^2}}{2}$

$BF=\dfrac{\sqrt{2(a^2+c^2)-b^2}}{2}$

$CG=\dfrac{\sqrt{2(a^2+b^2)-c^2}}{2}$

Also we know that $AD:DE=BD:DF=CD:DG=2:1$ , so:

$AD=\dfrac{\sqrt{2(b^2+c^2)-a^2}}{3}$

$BD=\dfrac{\sqrt{2(a^2+c^2)-b^2}}{3}$

$CD=\dfrac{\sqrt{2(a^2+b^2)-c^2}}{3}$

Finally, the value that we want is:

$\dfrac{2(b^2+c^2)-a^2+2(a^2+c^2)-b^2+2(a^2+b^2)-c^2}{9(a^2+b^2+c^2)}=\dfrac{3(a^2+b^2+c^2)}{9(a^2+b^2+c^2)}=\boxed{\dfrac{1}{3}}$

Let $m_a , m_b , m_c$ be the medians on $a = BC, b = AC, c = AB$ of $\triangle ABC$ . Using Stewart's theorem " $man+dad = bnb+cnc$ ", we obtain $\frac{a}{2}\cdot a\cdot\frac{a}{2}+m_a\cdot a\cdot m_a = b\cdot\frac{a}{2}\cdot b+c\cdot\frac{a}{2}\cdot c,$ $a^2+4m_a^2 = 2(b^2+c^2),$ $4m_a^2 = 2(b^2+c^2)-a^2.$ Adding this with two similar formulae for $m_b$ and $m_c$ , $4(m_a^2+m_b^2+m_c^2) = 3(a^2+b^2+c^2),$ $m_a^2+m_b^2+m_c^2 = \frac{3}{4}(a^2+b^2+c^2).$ Since D is the centroid of triangle $ABC$ , $AD = \frac{2}{3}\cdot m_a$ , or $AD^2 = \frac{4}{9}\cdot m_a^2$ . Adding this with two similar formulae for $BD^2$ and $CD^2$ , $AD^2+BD^2+CD^2 = \frac{4}{9}(m_a^2+m_b^2+m_c^2).$ It follows that $\frac{AD^2+BD^2+CD^2}{a^2+b^2+c^2}=\frac{AD^2+BD^2+CD^2}{m_a^2+m_b^2+m_c^2}\cdot\frac{m_a^2+m_b^2+m_c^2}{a^2+b^2+c^2}=\frac{4}{9}\cdot\frac{3}{4}=\frac{1}{3}.$

Little more general,

Let ABC be an equilateral triangle of side a .

Hence AD = BD = CD = a^2 by root 3 .

Well I don't know how to write that :(

numerator = a^2 denominator = 3 (a^2) hence numerator / denominator = 0.33

What is this?u're saying when 1:3=0.33333 to write as 0.34 and updated the answer as 0.333,please post answer correctly