Let $E$ be the foot from $D$ to $AB$ and let $F=DE\cap AC$. Because of the given condition we have $DF=FE$ and $AF=FC$. So $KF\perp AC$. We have $\angle AKF = 90 - \angle KAF = 90 - \angle CBA = 90 - \angle DAB = \angle ADE = \angle ADF \implies KDFA$ is cyclic. So $90 = \angle KFA = \angle KDA$.

Okay a new solution I think. A little harmonics is fine, right? I promise, I won't even project... Maybe this is just symmedians... Not even harmonics... Forgive me... Solution: Let $AB \cap \odot(ABCD) = D'$ and $E$ be the feet of perpendicular from $D$ to $AB$. Clearly, $DD'AC$ is a harmonic quadrilateral. [asy][asy] size(8cm); pair A = (-3,0); pair B = (3,0); pair C = (1,2); pair D = (-1,2); pair E = foot(D,A,B); pair O = circumcenter(A,B,C); pair D_ = 2*O - A; pair K = extension(O, bisectorpoint(A,C), D_,D); //yeah, kinda a hack, idk how to make tangents draw(circumcircle(A,B,C), orange); draw(A--B--C--D--A, blue); draw(A--K, red); draw(C--K, red); draw(K--D_, red); draw(D--E, blue); draw(A--D_, blue); draw(A--C, blue); draw(C--D_, blue); markscalefactor = 0.03; draw(rightanglemark(D,E,A), deepgreen); draw(rightanglemark(A,C,D_), deepgreen); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$D$", D, dir(D)); dot("$K$", K, dir(K)); dot("$D'$", D_, dir(D_)); dot("$E$", E, S); [/asy][/asy] By the length-condition of the trapezoid, we clearly have $\overline{AE} = \overline{CD}$. It is very to easy to see that $\measuredangle DAB = \measuredangle ABC = \measuredangle ACD'$. From the harmonic quadrilateral, we have \[\frac{\overline{CD'}}{\overline{AD'}} = \frac{\overline{CD}}{\overline{AD}} = \frac{\overline{EA}}{\overline{AD}}.\]By SAS similarity criteria, we can now conclude that $\triangle AED \sim \triangle D'CA$. Finally \[ \angle KDA = 180^\circ - \angle ADD' = 180^\circ - \angle ACD' = 180^\circ - 90^\circ = 90^\circ\]and we're done! $\blacksquare$