Reflect $A$ over $M$ to get $A'$, and let $E$ be the second point of intersection of the mentioned circumcircle and $AB$. Then we have $TAA'E$ cyclic because $DE \parallel BC \parallel TA \parallel A'D$ and $AE = AD = TA'$. Also note that $\angle ATK = 180^\circ - \angle KCB = \angle KEB$, so $A, T, A', K$ are cyclic. From this we have $\angle TKA = \angle TA'A = \angle MAC$, so we are done.

Lemma: if we have $\frac{sin(\alpha)}{sin(k-\alpha)} = \frac{sin(\beta)}{sin(k-\beta)}$ then we have $\alpha = \beta$ or $ k = 180$ Main problem: Let $X$ be the intersection of the circle of triangle $BDC$ with $AB$. we have $\angle XKT = \angle ABC = \angle ACB = \angle CAT$ and we want to prove the equality of $\angle AKT$ and $\angle CAM$ and we have $\angle TCB = \angle KXA = 180 - (\angle ATK)$ thus we have $sin(\angle ATK) = sin(\angle AXK)$ . because of the lemma, we have : $$ \frac{sin(\angle AKT)}{sin(\angle AKZ)} = \frac {\frac{ AT . sin(\angle ATK)}{AK}}{ \frac{AX .sin(\angle AXK)}{AK}} = \frac{AT}{AX} = \frac{AT}{AD} = \frac{ sin(\angle TAM)}{sin( \angle DAM)}$$ thus we have $\angle AKT = \angle CAM$. DONE!

Good Problem $ \angle CAM = \angle CAK + \angle KAM $ and also $ \angle AKT = \angle CAK + \angle ACK $ thus it suffices to prove $ \angle KAM = \angle ACK $ Reflect $A$ Over $M$ to get point $P$ so the quadrilateral $ATPD$ is a parallelogram and we know that $BDKC$ is a cyclic quadrilateral therefore $\angle DBK = \angle DCK$ and because $ AC \parallel TP $ so $\angle ACT = \angle PTC$ therefore in order to show $ \angle KAM = \angle ACK $ it suffices to prove $ATPK$ is cyclic which means $ \angle ATK = \angle APK $ and $\angle PAK = \angle PTK$ , and we knew $ \angle ATK = \angle KDB $ and $\angle PTK = \angle DBK$ therefore it’s enough to show that the triangles $\triangle KBD$ and $\triangle KAP$ are similar which by spiral similarity is equivalent to $KPD$ and $KAB$ being similar and we knew $DP \parallel BC$ therefore $\angle PDC = \angle BCD$ therefore $ \angle PDK = \angle ACB - \angle KBC $ therefore $ \angle ABK = \angle PDK = \angle ACB - \angle KBC $ therefore it suffices to prove $ \frac{AB}{DP} = \frac{BK}{DK} $ and we said beforehand that $ATPD$ is a parallelogram so $DP = AT$ , therefore we should show $\frac{AB}{AT} = \frac{BK}{DK}$ and its easy to see that $\triangle DBK$ and $\triangle TCA$ are similar so $\frac{AC}{AT} = \frac{BK}{DK}$ and we knew $AC = AB$ , so $ \frac{AB}{DP} = \frac{BK}{DK} $ and the problem is solved .

Let $A'$ be the reflection of $A$ over $M$ and let $E = (BCD)\cap AB$. It is clear that $EATA'$ is an isosceles trapezoid and thus it its cyclic. It is sufficient to show that $K$ lies on that circle. But that is true because $\angle TA'E = \angle AED = \angle ABC = \angle TKE$. $\square$

Let $A'$ be a point such that $ADA'T$ is parallelogram. Let $E = (AKT) \cap AB$ with $E \neq A$. Note that: $AEA'T$ is an isosceles trapezoid. Notice that: $$\angle AKT = \angle AA'T = \angle CAA' = \angle CAM.$$