KöMaL 2020 B5080

Let $D$ denote the midpoint of the base $AB$ in an isosceles triangle $ABC$. Let $H$ be the point lying closer to $C$ that divides the leg $AC$ in a $1:2$ ratio. The circle $BCH$ intersects line $CD$ at the points $C$ and $X$. Show that $CX=\frac 43r$, where $r$ is the radius of the circle $ABC$.