Solving Triangles - Part 4

Let $C$ be at the origin, $B$ be at $B(4, 0)$ , and $A$ be at $A(4, 2\sqrt{3})$ . Let $D_1$ be a point on the left side of $AB$ . Since $\angle ADB = 120°$ , and the inscribed angle is half the central angle, the locus of points of $D_1$ is a $120°$ arc of circle $E$ passing through $A$ and $B$ .

With some trigonometry, the center of circle $E$ is $E(5, \sqrt{3})$ with a radius of $2$ . Since $D_1$ is nearest to $C$ along the line $CE$ , which by the distance formula is $CE = \sqrt{5^2 + \sqrt{3}^2} = 2\sqrt{7}$ , and since circle $E$ has a radius of $2$ , $CD_1 = 2\sqrt{7} - 2$ .

Let $D_2$ be a point on the right side of $AB$ . Then by a similar argument, the locus of points of $D_2$ is a $120°$ arc of circle $F$ passing through $A$ and $B$ , but this time at center $F(3, \sqrt{3})$ . Since $D_2$ is furthest from $C$ along the line $CF$ , which by the distance formula is $CF = \sqrt{3^2 + \sqrt{3}^2} = 2\sqrt{3}$ , and since circle $F$ has a radius of $2$ , $CD_2 = 2\sqrt{3} + 2$ .

Therefore, the range of $CD$ is $\boxed{[2\sqrt{7} - 2, 2\sqrt{3} + 2]}$ .