Let $O$ be the circumcenter of the triangle $\triangle ABC$. Let $\angle BAC = \alpha$, $\angle ABC = \beta$, $\angle ACB = \gamma$. We construct the line passing through $D$ and $O$. Let $L$ be the point of intersection of this line and the ray $BI$. Since $AI$ is a bisector of the angle $BAC$, we have $BD = DC$, and so the line $DO$ is a perpendicular bisector of the segment $BC$. Therefore, $BL = CL$, i.e. the triangle $BLC$ is isosceles and $\angle LBC = \angle LCB = \beta/2$ ($BI$ is the bisector of the angle $ABC$). Hence,

$$\angle ILC = \angle BLC = 180^{\circ} - (\angle LBC + \angle LCB) = 180^{\circ} - \beta.$$

On the other hand,
$$\begin{align*} \angle CDI &= \angle CDA = 180^{\circ} - (\angle DAC + \angle ACD) = \\ &= [\angle DCB = \angle DAB = \angle DAC = 0.5\alpha, \angle ACB = \gamma, \angle ACD = \\ &= \angle ACB + \angle DCB = \gamma + 0.5\alpha] = 180^{\circ} - (0.5\alpha + \gamma + 0.5\alpha) = 180^{\circ} - \alpha - \gamma = \beta. \end{align*}$$

Therefore, $\angle ILC + \angle CDI = 180^{\circ}$. It follows that the points $I, D, C$, and $L$ lie on the same circumference (passing through $I, D, C$). Thus, $K$ and $L$ coincide. Hence, $CK = CL = BL = BK$, as required.