Circling The Foot

Construct the point $Y$ on $XB$ extended such that $BY$ = $XA$ . Clearly $CAXB$ is cyclic, so we have $\angle CAX = 180^\circ - \angle CBX = \angle CBY$ . Thus triangles $CAX$ and $CBY$ are congruent by side-angle-side (with $CA = CB$ , $XA = BY$ and $\angle CAX = \angle CBY$ ). Therefore $CX = CY$ and this implies that $CXY$ is an isosceles triangle. Since $D$ is the foot of the perpendicular from $C$ to $XB$ , $D$ is the midpoint of $XY$ . So $XD = \frac{XY}{2} = \frac{XB + BY}{2} = \frac{300 + 100}{2} = 200$ .