We just aply the median formula and sum the 3 relations.Why does it have to be right angled?The relation is true in any triangle.

hello, we get $m_1^2=s_2^2+\frac{s_1^2}{4}$ $m_2^2=s_1^2+\frac{s_2^2}{4}$ $m_3^2=\frac{s_3^2}{4}$ thus we get $m_1^2+m_2^2+m_3^2=\frac{3}{4}s_1^2+\frac{3}{4}s_2^2+s_1^2+s_2^2+\frac{s_3^2}{4}=$ $\frac{3}{4}(s_1^2+s_2^2+s_3^2)$ Sonnhard.

hello, in every triangle we get $a^2+b^2+c^2=\frac{3}{4}(m_a^2+m_b^2+m_c^2)$ Sonnhard.

http://en.wikipedia.org/wiki/Apollonius'_theorem

Let BC=s1 with median m1 AC=s2 with median m2, AB=s3 with median m3 So, s2 ²= s1²+s3² m1²=s3²+(s1²)/4 m3²=(s3²)/4+s1² m1²+m3²=5/4{s1²+s3²} By Stewart's theorem, (m2)²*s2+s2³/4 =(s2/2){s1²+s3²} (m2)²=(s2²)/4 (m2)²=(s1²+s3²)/4 m1²+m3²+m2²=(s1²+s3²)/4+(5/4){s1²+s3²} = (3/2){s1²+s3²} =(3/4){2s1²+2s3²} =(3/4){ s1²+s3²+s1²+s3²} =(3/4){ s1²+s2²+s3²}(Proved) @Krishijivi