Let $G=OF\cap BC$. Claim: $OECG$ is cyclic. First note that $DC^2=DA\cdot DB= DE\cdot DO$ but $\angle DCO=90^{\circ}$ meaning $OEC=90^{\circ}$. Now $\angle OBC+\angle BOF= \angle OBC+ \angle BAC=90^{\circ}-\frac{1}{2} \angle BOC+ \angle BAC=90^{\circ}$ therefore $\angle OGC=90^{\circ}$ . Hence $OECG$ is cyclic. Claim: $TB=TC$. Since $OB=OC$ We have $G$ is the midpoint of $BC$ and $T\in OG$ so $TB=TC$. We need $\angle TOK= \angle TBK$ but $\angle TOK = \angle GOE= 18 0^{\circ}-\angle ECG= 180^{\circ}-\angle ABC=TBK$ so we're done.

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An alternative solution: Let $X$ be the reflection of $C$ over $OD$. If $E'$ is the midpoint of $XC$ on $OD$, then \[\vert DE'\vert \cdot \vert DO\vert=\vert DC\vert^2=\vert DB\vert \cdot \vert DA,\]hence $ABE'O$ is cyclic and hence $E=E'$. We claim that $X$ also lies on $(OBTK)$. Indeed, \[\angle KOT=\angle ECK=\angle KXT,\]so $KTOX$ is cyclic. Moreover, $O$ is the $K$-northpole of $XKB$, hence also $KBOX$ is cyclic. Done!