Let the midpoint of $BC$ is $M$ and let the circumcircles of $\bigtriangleup BTM$ and $\bigtriangleup DCT$ intersect at for a second time at $F\neq T$. $KC$ is tangent to $(ABC)$ since $\angle TAC=\angle TDK=\angle TCK$. $K$, $M$, $F$ are colinear because $\angle TFK=\angle TDK=\angle TAD=\angle TAC=\angle TBC=\angle TBM=\angle TFM$. $AB$ is tangent to $(BTMF)$ since $\angle BTM=\angle ATC=\angle ABC$. $B$, $F$, $D$ are colinear since $\angle DFK=180^{\circ}-\angle ACK=\angle ABC= \angle BTM=\angle180^{\circ}-BFK$. Now we have $\angle KMC=\angle FMB=\angle ABD=180^{\circ}-2\angle A$ and $\angle MCK=\angle A$ so $\bigtriangleup MCK$ is isosceles. Hence $K$ belongs to the circle with diameter $BC$, i.e. $\angle BKC=90^{\circ}$.

Hello,this is my solution.

Let $M$ be the midpoint of $BC$. Note that : $KDT=TAD=TAC=MAB$ , also :$TKD=TCD=C+BAT=C+CAM=AMB$ So $\triangle DKT \sim \triangle AMB$ which means $TD/TK=AB/BM$. Also , $AB/DT=AM/DK$ . But $BD/DT=AB/DT$. But it's enough to show that $MK=MB$ which is $BM/TK=MK/TK$. But we had $TDK=TAC=TBM$ so it's enough to show that $\triangle TBM \sim \triangle TDK$ and $T$ would be the center of spiral similarity sending $BM$ to $KD$. But we have : $DTK=ABC$ and also , $MTB=MXB=ABC$ where $X$ is A-Humpty point. So $DTK=MTB$.So $T$ is indeed the center of spiral similarity sending $BM$ to $KD$. Finally we have : $TB/BD=TM/MK$ and we have : $TB/TM=AB/BM=BD/BM$ .So $BM=MK$ and we're done

Let $M$ be midpoint of $BC$.It suffices to show that $BM=MK$ Claim1: $\triangle ABM \sim \triangle DTK$ $\angle BAM=\angle CAT=\angle DAT =\angle TDK$ $\angle BMA=\angle MAC+\angle MCA=\angle BAT+\angle MCA=\angle BCT+\angle MCA=\angle TCD=\angle TKD \ \ \ \square$ Claim2: $\triangle ABM \sim \triangle DTK \sim \triangle BTM$ It is well known symmedian lemma that $BC$ is angle bisector of $\angle AMT \implies \angle AMB=\angle BMT$ Also we have $\angle MAB =\angle CAT=\angle CBT=\angle MBT \ \ \ \square$ Claim3: $\triangle BDT \sim \triangle MKT$ By Claim2 $T$ is center of spiral similarity sends $BM$ to $DK$.So $T$ is also center of spiral similarity sends $BD$ to $MK \ \ \ \square$ Now by Claim2 we have $ \frac{AB}{BM}=\frac{BT}{TM} \iff \frac{BD}{BM}=\frac{BT}{TM}$ And by Claim3 we have $ \frac{BD}{MK}=\frac{BT}{TM}$ Thus $\frac{BD}{BM}=\frac{BD}{MK} \implies BM=MK$ So we are done

Let $X$ = $\overleftrightarrow{BB} \cap \overleftrightarrow{CC}$ ( wrt. $\odot(ABC)$ ) , $\angle{BAC} = \alpha$ , $\angle{CBA} = \beta$ , $\angle{ACB} = \gamma$ and $\angle{TAC} = \theta$ it is easy to prove that $\overleftrightarrow{DD}$ (wrt. $\odot(ADT)$ ) , $\odot(DCT)$ and $\overline{CX}$ met at the same point Assume $K'$ be point on $\overline{CX}$ such that $\overline{BK'} \bot \overline{CX}$ , it suffices to show that $K'$ $\in$ $\odot(CDT)$ Claim : $\triangle$ $TCK'$ $\sim$ $\triangle$ $TAD$ Proof : Since $\angle{TCK'} = \theta = \angle{TAD}$ , it suffices to show that \begin{align*} \frac{CK'}{TC} &= \frac{AD}{AT} \\ \frac{BCcos\alpha}{BC \times \frac{sin\theta}{sin(180-\alpha)}} &= \frac{AB \times \frac{sin2\alpha}{sin\alpha}}{AB \times \frac{sin(\alpha + \gamma - \theta)}{sin \gamma}} \\ \frac{sin\alpha cos\alpha}{sin \theta} &= \frac{2cos\alpha sin\gamma}{sin(\beta+\theta)} \\ \frac{sin\alpha}{sin\theta} &= \frac{2sin\gamma}{sin(\beta+\theta)} \\ \frac{sin(\beta+\theta)}{sin\theta} &= \frac{2sin\gamma}{sin\alpha} \\ \frac{AT}{TC} &= \frac{2AB}{BC} \end{align*} Let $P$ be reflect of $B$ over $A$ Since -1 = $(A,T ; B,C)$ $\rightarrow$ $TC \times AB$ = $BT \times AC$ $\rightarrow$ $TC \times PA$ = $BT \times AC$ $\rightarrow$ $\frac{PA}{BT} = \frac{AC}{TC}$ and $\angle{PAC} = \angle{BTC}$ then $\triangle$ $PAC$ $\sim$ $\triangle$ $BTC$ Spiral similarity implies that $\triangle$ $CTA$ $\sim$ $\triangle$ $CBP$ $\rightarrow$ $$\frac{AT}{TC} = \frac{PB}{BC} = \frac{2AB}{BC}$$Thus the claim is proven $\square$ By Claim implies that $\angle{TK'C} = \angle{TDA}$ Hence $K'$ $\in$ $\odot(CDT)$ which solved the problem $\blacksquare$