$ABCD$ is a convex quadrangle, $G$ is its center of gravity as a homogeneous plate (i.e., the intersection point of two lines, each of which connects the centroids of triangles having a common diagonal). a) Suppose that around $ABCD$ we can circumscribe a circle centered on $O$. We define $H$ similarly to $G$, taking orthocenters instead of centroids. Then the points of $H, G, O$ lie on the same line and $HG: GO = 2: 1$. b) Suppose that in $ABCD$ we can inscribe a circle centered on $I$. The Nagel point N of the circumscribed quadrangle is the intersection point of two lines, each of which passes through points on opposite sides of the quadrangle that are symmetric to the tangent points of the inscribed circle relative to the midpoints of the sides. (These lines divide the perimeter of the quadrangle in half). Then $N, G, I$ lie on one straight line, with $NG: GI = 2: 1$.
Problem
Source: Sharygin 2006 CR 23 (a) IX-X (b) X-XI
Tags: orthocenter, Centroid, Nagel point, cyclic quadrilateral, tangential quadrilateral, geometry