In $$\triangle ABC$$, the medians through $$B$$ and $$C$$ are perpendicular. Then $$b^2+c^2$$ is equal to

Let the vectors from the centroid to the vertices be $$\mathbf{a},\mathbf{b},\mathbf{c}$$, so $$\mathbf{a}+\mathbf{b}+\mathbf{c}=0$$. The median through $$B$$ is parallel to $$\mathbf{b}$$ and the median through $$C$$ is parallel to $$\mathbf{c}$$, so perpendicular medians give $$\mathbf{b}\cdot\mathbf{c}=0$$. Since $$\mathbf{a}=-(\mathbf{b}+\mathbf{c})$$, we get $$a^2=b^2+c^2+2\mathbf{b}\cdot\mathbf{c}$$ in centroid-vector lengths; converting these vector lengths to the corresponding side lengths gives $$b^2+c^2=5a^2$$.