Let $E, F$ be intersections of $KD$ and $AC$, $KB$ and $DC$. We will prove that triangle $KCE$ is isosceles at C. $\frac{CE}{BD} = \frac{AC}{AB}$ Note that $\angle KDC = \angle ABC$, we will try to prove triangles $ABC$ and $CDK$ are similar. $\frac{KD}{BC} = \frac{DF}{CF} = \frac{DC}{AC}$ using Menelaus theorem and so triangles $ABC$ and $CDK$ are similar. Therefore $\frac{CE}{BD} = \frac{CK}{CD}$, yeilds $CE=CK$. Now using angles-chasing and the conclusion follows.

Let the parallel to \(AC\) through \(K\) intersect \(AB\) at \(A'\) and \(BC\) at \(C'\). We easily get that \(K\) is the midpoint of \(A'C'\). From there, we get that \(DK\) is a midline of \(A'BC'\), and as such, \(DA' = DB = DC\). This, along with \(\angle A'DK = \angle DBC = \angle DCB = \angle CDK\) implies that \(C\) is the reflection of \(A'\) across \(DK\). This implies \(\angle DCK = \angle DA'K = \angle DAC\), as desired.