Assume w.l.o.g. that $\angle ABC > \angle ACB$ (Fig. 31; otherwise change the roles of points $B$ and $C$). Note that
$$\angle ADK = 180^{\circ} - \angle CDA = \angle DAC + \angle ACD = \angle BAD + \angle ACB.$$
By inscribed angle property, $\angle KAB = \angle ACB$, whence
$$\angle BAD + \angle ACB = \angle BAD + \angle KAB = \angle KAD.$$
Consequently, $\angle ADK = \angle KAD$, which implies $KA = KD$.

Let $O$ be the circumcenter of the triangle $ABC$, $M$ be the midpoint of the side $BC$, and $X$ be the point of intersection of the ray $OM$ with the circumcircle of the triangle $ABC$. Moreover, let $Q$ be the point of intersection of the line perpendicular to the side $BC$ and passing through point $D$ with the line $AO$ (Fig. 32). As $OM$ is the perpendicular bisector of the side $BC$, point $X$ bisects the arc $BC$ of the circumcircle of the triangle $ABC$, whence the line $AD$ also passes through $X$. As $OA = OX$, we have $\angle XAO = \angle OXA$. On the other hand, $OX \perp BC$ and $QD \perp BC$ together imply $OX \parallel QD$, whence $\angle OXA = \angle QDA$. Thus $\angle DAQ = \angle XAO = \angle QDA$, implying $AQ = QD$. Consequently, there exists a circle with center $Q$ passing through both points $A$ and $D$. As $KA \perp AQ$ and $KD \perp DQ$, the lines $KA$ and $KD$ are tangent to this circle. By the property of tangent line segments, $KA = KD$.