Estonian Math Competitions

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 $KA=KD$.

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Alternative solution.

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, $KA=KD$.