Ukrainians really like those angle condition problems, huh? WLOG, $\angle ABC\geq \angle ACB$. Construct points $E$ and $F$ such that $\triangle EBL\sim\triangle KLC$ and $\triangle FCL\sim\triangle KLB$. Note that $P$ is the second intersection of $(EBL)$ and $(FCL)$. Also let $Q=(EBLP)\cap \overline{AB}$ and $R=(FCLP)\cap \overline{AC}$. Let $N$ be the reflection of $L$ over $D$. Observe that $P$ lies on $(AQR)$ by Miquel theorem. [asy][asy]import olympiad; size(8cm);defaultpen(fontsize(10pt));pen org=magenta;pen med=mediummagenta;pen light=pink;pen deep=deepmagenta; pair A,B,C,I,L,K,D,E,F,P,M,N,Q,R; A=dir(130);B=dir(205);C=dir(335);I=incenter(A,B,C);L=extension(A,I,B,C); K=0.3A+0.7L; E=(K-L)/(L-C)*(B-L)+B;F=(K-L)/(L-B)*(C-L)+C;D=foot(K,B,C);P=intersectionpoints(circumcircle(E,B,L),circumcircle(C,L,F))[0]; M=midpoint(K--P);N=2D-L;Q=intersectionpoints(circumcircle(E,B,L),A--B)[0]; R=intersectionpoints(circumcircle(C,F,L),A--C)[0]; draw(A--B--C--cycle,heavygreen);draw(circumcircle(C,F,L),deepcyan);draw(circumcircle(B,E,L),deepcyan);draw(circumcircle(A,R,Q),deepcyan); draw(B--E--F--C,heavygreen);draw(A--P--L--cycle,cmyk(red)+1);draw(K--N--A--cycle,cmyk(blue)+1);draw(Q--R,heavygreen);draw(A--M--D--cycle,magenta+1);draw(E--L--F,heavygreen); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$K$",K,dir(90)); dot("$L$",L,dir(L)); dot("$E$",E,dir(E)); dot("$F$",F,dir(F)); dot("$P$",P,dir(90)); dot("$D$",D,dir(D)); dot("$N$",N,dir(N)); dot("$M$",M,dir(90)); dot("$Q$",Q,dir(Q)); dot("$R$",R,dir(R)); [/asy][/asy] Claim: $E,K,P,F$ are collinear. Proof. Collinearity of $P,E$ and $F$ comes from the angle chase; \begin{align*} \measuredangle EPL=\measuredangle EBL=\measuredangle KLC=\measuredangle FCL=\measuredangle FPL, \end{align*}where we used the fact that $AL\parallel BE\parallel CF$. Collinearity of $K,E$ and $F$ comes from the similarity; \begin{align*} \frac{KL}{EB}=\frac{LC}{BL}=\frac{FC}{KL},\end{align*}which implies the collinearity along with $KL\parallel BE\parallel CF$ (this is a well-known lemma). Claim: $\triangle ARK\sim\triangle ALB$. Proof. Firstly, $K$ lies on $(AQRP)$ as $\measuredangle KPR=\measuredangle FPR=\measuredangle CFR=\measuredangle KAR$. Now, since $\measuredangle ARL=\measuredangle CRL=\measuredangle CPL=\measuredangle LKB=\measuredangle AKB$ and $\measuredangle BAK=LAR$, we conclude that $\triangle ABK\sim\triangle ALR$, hence $A$ is the center of spiral similarity sending $\overline{BK}$ to $\overline{LR}$, ergo $A$ is also the center of spiral similarity sending $\overline{BL}$ to $\overline{KR}$, the claim follows. Therefore, $\measuredangle APK=\measuredangle ARK=\measuredangle ALB=\measuredangle FCL=\measuredangle FPL=\measuredangle KPL\implies \overline{KP}$ bisects $\angle APL$. Hence, $$\frac{AP}{PL}=\frac{AK}{KL}=\frac{AK}{KN}.$$On top of that $\measuredangle AKN=\measuredangle LKN=2\measuredangle ALB=2\measuredangle APK=\measuredangle APL$, hence we conclude that $\triangle AKN\sim\triangle APL$. By gliding principle as $M,D$ are the midpoints of $\overline{KP},\overline{NL}$, respectively, $\triangle AKN\sim\triangle APL\sim\triangle AMD$. Finally, \begin{align*} \measuredangle AMD=\measuredangle APL=2\measuredangle ALB=\measuredangle ACB+\measuredangle ABC. \end{align*}We are done.

Let $S$ be the intersection of the tangent to $ABC$ at $A$. We need to prove $AMDS$ is cyclic by angles. Trivially $BPKC$ is cyclic. To prove $S,P,K$ are collinear, use a phantom point. Let $K' = SP \cap BPC$. Not hard by a few angles to see $\angle SLP = \angle SK'L$ which implies $K' \equiv K$ (probably could've been done cleaner with radicals). Now let $T$ be the circumcenter of $APK$. $\angle AKS = \angle ALS = \angle SAP$ we have $ATMS$ cyclic and $\angle TPA = \angle DPL$ hence $ATDS$ cyclic so $ATMDS$ cyclic which means $\angle AMD = 180^{\circ} - \angle ASD$ which is what we wanted to prove. (for motivation extend $PL,PK$ to the circle $BPC$)

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