Prove that there exists a point $K$ in the plane of $\vartriangle ABC$ such that $$AK^2 - BC^2 = BK^2 - AC^2 = CK^2 - AB^2.$$Let $Q, N, T$ be the points of intersection of the medians of the triangles $BKC, CKA, AKB$, respectively. Prove that the segments $AQ, BN$ and $CT$ are equal and have a common point.