Here's my solution: Let $AK \cap \odot (BCD) =X$. Then $\angle AKT=\angle CKX=\angle CDX$. So showing $\angle AKT=\angle CAM$ is equivalent to proving that $AM \parallel DX$. Now, animate $K$ on $\odot (BCD) $, keeping $D$ as fixed. Then $K \mapsto T$ is projective (perspectivity at $C$). Also homothety at $D$ with radius $0.5$ gives that $T \mapsto M$ is also a projective map. Finally we have the following sequence of projective maps, using various sequences of perspectivities:- $$M \mapsto AM \mapsto \infty_{AM} \mapsto D\infty_{AM} \cap \odot (BCD) $$And $K \mapsto X$ is also a projective map (in fact a perspectivity at $A$). So it suffices to a show that $DX \parallel AM$ for three positions of $K$. The result is obvious when $K=B, D$ (these give degenerate cases). For the third position take $K=AB \cap \odot (BCD) $ or $K=C$. Then it's not difficult to prove the desired statement (I would have done so, but it would take too much time to write, and I have to go play TT ). Hence, done. $\blacksquare$

When you realize that that you've found out a solution that the person exactly above you has posted.... This is one of my beginning attempts at using moving points, so please notify if I made mistakes.

Let $N$ be the reflection of $A$ over $M$, and $E$ be the intersection point of $\odot(BCD)$ with $AB.$ Then $ADNT$ is a parallelogram and $AT \parallel BC \parallel DE$, meaning that $E$ is on $ DN$. Now, $\angle{EAT}=180^{\circ}-\angle{B}=180^{\circ}-\angle{TKE} \Longrightarrow ATKE$ is cyclic. Also, $AT \parallel NE$, $AE=AD=TN$, which means that $ATNE$ is an isosceles trapezoid, that is, cyclic. So $ATNKE$ is also cyclic. Now, $\angle{CAM}=\angle{DAM}=\angle{ANT}=\angle{AKT}$.

Let $O_2$ be the circumcenter of $\triangle AKC$, it suffices to prove that $MA$ is tangent to the circumcircle of $\triangle AKC$, which is $MA^2 = pow(M, O_2)$. By Steward's thm, we have \begin{align*} MA^2 & = \frac{2AD^2+2AT^2 -DT^2}{4} \\ pow(M, O_2) &= MO_2^2 - R^2 = \frac{2DO_2^2+2TO_2^2 -DT^2}{4}-R^2 \\ &= \frac{2pow(D,O_2)+2pow(T,O_2) -DT^2}{4}. \end{align*}Moreover, let $AT \cap \odot O_2 = Q$, then $pow(D,O_2) = -DA \cdot DC$ and $pow(T,O_2) = TA \cdot TQ$. Thus it suffices to prove \[ AD^2+ AT^2 = -DA \cdot DC+ TA \cdot TQ \Leftrightarrow AD \cdot AC = TA \cdot TQ \]Finally, by basic angle chasing, we can prove that $TD$ intersect $QB$ on the circle $\odot BCD$. Again by basic angle chasing we have $\triangle ADT \sim \triangle AQB$, which leads to the desired equation.

[asy][asy] size(9cm); defaultpen(fontsize(10pt)); pair B = dir(230), C = dir(310), D = dir(75), E = dir(105), A = extension(B,E,C,D), K = dir(355), T = extension(A,A+dir(B--C),C,K), M = (D+T)/2, P = 2M-A; dot("$A$", A, dir(135)); dot("$B$", B, dir(230)); dot("$C$", C, dir(310)); dot("$D$", D, dir(225)); dot("$E$", E, dir(140)); dot("$K$", K, dir(315)); dot("$T$", T, dir(45)); dot("$M$", M, dir(90)); dot("$P$", P, dir(345)); draw(unitcircle^^circumcircle(A,E,K), dotted); draw(M--D--A--M--T--A--B--C--T^^E--K--A--M^^C--D); draw(E--P--T^^P--M, dashed); [/asy][/asy] Let $E$ be the second intersection point of $\overline{AB}$ and $(BCD)$, and $P$ be the reflection of $A$ over $M$. Claim: Points $A$, $T$, $P$, $K$, $E$ are concyclic. Proof. First we show that $AEKT$ is cyclic by angle chase: $$\angle TAE = 180^\circ - \angle ABC = 180^\circ - \angle CKE.$$Note that $\overline{DE} \parallel \overline{BC} \parallel \overline{AT}$, and $ADPT$ is a parallelogram by the way we defined $P$, so $E, D, P$ are collinear. Then $$\angle TPE = \angle TAD = \angle ADE = \angle AED,$$so $AEPT$ is an isosceles trapezoid and the conclusion follows. $\square$ We finish the problem with some more angle chase: $$\angle AKT = 180^\circ - \angle AKE - \angle EKC = 180^\circ - \angle APE - \angle EDC = 180^\circ - \angle APE - \angle ADP = \angle CAM,$$and we are done. $\blacksquare$

Here is an unconventional trig approach for the problem. Let $E\not = B$ be the intersection of $(BCD)$ and $AB$. Claim: $A,T,E,K$ concyclic. Proof: Since $AT\parallel BC$, we have \[ \angle ATK=180-\angle TCB=180-\angle KCB=\angle BEK=180-\angle AEK,\]as desired. $\square$ We need to show $\angle DAM=\angle AET$. By the ratio lemma, we have \[ 1=\frac{DM}{MT} =\frac{AD}{AT} \frac{\sin \angle DAM}{\sin \angle MAT} = \frac{AE}{AT} \frac{\sin \angle DAM}{\sin \angle MAT} \implies \frac{AE}{AT} = \frac{\sin \angle MAT}{\sin \angle DAM}. \]But by the Sine Law, \[ \frac{AE}{AT} = \frac{\sin \angle ATE}{\sin \angle AET} \implies \frac{\sin \angle MAT}{\sin \angle DAM} = \frac{\sin \angle ATE}{\sin \angle AET}. \]Here is the key step. Notice that \begin{align*} \angle ATE+\angle AET &= 180-\angle BAT=\angle ABC\\ \angle MAT+\angle DAM &= \angle TAD = \angle TAC=\angle ACB=\angle ABC. \end{align*}So we actually have \[ \frac{\sin (\angle ABC-\angle DAM)}{\angle DAM} = \frac{\sin (\angle ABC-\angle AET)}{\sin \angle AET}.\]But if $\alpha+\beta=C$ for a constant $C \le \pi/2$, then $\tfrac{\sin \alpha}{\sin \beta}$ increases as we increase $\alpha$ since $\sin \alpha$ increases and $\sin \beta$ decreases. Hence $\tfrac{\sin \alpha}{\sin \beta}$ is injective over this range. Therefore, $\angle DAM=\angle AET$, and we are done.

Here is a video of my solution on my Youtube channel, which is essentially the same as both fastlikearabbit and MP8148's solution. https://www.youtube.com/watch?v=wgMTuWkIYcw

Cute. Let $(BDC)$ meet $AB$ at $E$. Since $\angle AEK = 180 - \angle BEK = \angle BCK = 180 - \angle KTA$ and so $E \in (ATK)$ Let $A'$ be the point such that $ADA'T$ is a parallelogram. Note that $A'$ lies on $DE$ and since $AE = AD = TA'$, we have that $ATA'E$ is an isosceles trapezoid $\implies A' \in (ATK)$ So, $\angle CAM = \angle AA'T = \angle AKT$, as desired. $\blacksquare$

XbenX wrote: In the segment $AC$ of an isosceles triangle $\triangle ABC$ with base $BC$ is chosen a point $D$. On the smaller arc $CD$ of the circumcircle of $\triangle BCD$ is chosen a point $K$. Line $CK$ intersects the line through $A$ parallel to $BC$ at $T$. $M$ is the midpoint of segment $DT$. Prove that $\angle AKT=\angle CAM$. (A.Kuznetsov) [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(12cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -7.266794776780934, xmax = 14.39240327147833, ymin = -6.839395137785745, ymax = 6.537042102652383; /* image dimensions */ pen qqwawe = rgb(0,0.41568627450980394,0.43137254901960786); pen qqttqq = rgb(0,0.2,0); pen qqzzcc = rgb(0,0.6,0.8); pen cczzqq = rgb(0.8,0.6,0); pen wwccqq = rgb(0.4,0.8,0); pen qqttzz = rgb(0,0.2,0.6); pen qqwwzz = rgb(0,0.4,0.6); pen zzwwqq = rgb(0.6,0.4,0); pen wwzzqq = rgb(0.4,0.6,0); pen ffqqtt = rgb(1,0,0.2); pen ffdxqq = rgb(1,0.8431372549019608,0); pen bfqqtt = rgb(0.7490196078431373,0,0.2); draw((1.84722990703285,4.165817591005137)--(-1.38666,-2.59007)--(4.64606,-2.7816)--cycle, linewidth(0.4) + qqwawe); /* draw figures */ draw((1.84722990703285,4.165817591005137)--(-1.38666,-2.59007), linewidth(0.4) + qqwawe); draw((-1.38666,-2.59007)--(4.64606,-2.7816), linewidth(0.4) + qqwawe); draw((4.64606,-2.7816)--(1.84722990703285,4.165817591005137), linewidth(0.4) + qqwawe); draw(circle((1.6829084436362738,-1.009900983953319), 3.452417203850304), linewidth(0.8) + qqttqq); draw((1.84722990703285,4.165817591005137)--(4.968572795803814,0.050096334126780384), linewidth(0.4) + qqzzcc); draw((4.00738903698368,3.1866817300171415)--(1.84722990703285,4.165817591005137), linewidth(0.4) + cczzqq); draw((4.00738903698368,3.1866817300171415)--(4.968572795803814,0.050096334126780384), linewidth(0.4) + wwccqq); draw((4.968572795803814,0.050096334126780384)--(2.5903848294873786,2.3211155513650166), linewidth(0.4) + qqttzz); draw((0.9885564173035511,2.3719712509144015)--(6.16754816693451,2.207545869029146), linewidth(0.4) + qqwwzz); draw((6.16754816693451,2.207545869029146)--(4.00738903698368,3.1866817300171415), linewidth(0.4) + zzwwqq); draw((6.16754816693451,2.207545869029146)--(5.424393244479982,4.052247908669266), linewidth(0.8) + wwzzqq); draw(circle((3.576335914190062,2.2356969118767034), 2.5913651653108176), linewidth(0.4) + ffqqtt); draw((1.84722990703285,4.165817591005137)--(5.424393244479982,4.052247908669266), linewidth(0.4) + ffdxqq); draw((5.424393244479982,4.052247908669266)--(4.64606,-2.7816), linewidth(0.4) + ffqqtt); draw((2.5903848294873786,2.3211155513650166)--(5.424393244479982,4.052247908669266), linewidth(0.4) + bfqqtt); /* dots and labels */ dot((4.64606,-2.7816),linewidth(4pt) + dotstyle); label("$C$", (4.73656446877788,-2.609429056786271), NE * labelscalefactor); dot((-1.38666,-2.59007),linewidth(4pt) + dotstyle); label("$B$", (-1.2872615732737753,-2.410111283336034), NE * labelscalefactor); dot((1.84722990703285,4.165817591005137),linewidth(4pt) + dotstyle); label("$A$", (1.9461156404745399,4.344546594699776), NE * labelscalefactor); dot((2.5903848294873786,2.3211155513650166),linewidth(4pt) + dotstyle); label("$D$", (2.676947476458748,2.506393795103146), NE * labelscalefactor); dot((4.968572795803814,0.050096334126780384),linewidth(4pt) + dotstyle); label("$K$", (5.046614338589363,0.22531261006154446), NE * labelscalefactor); dot((5.424393244479982,4.052247908669266),linewidth(4pt) + dotstyle); label("$T$", (5.511689143306586,4.233814498338534), NE * labelscalefactor); dot((4.00738903698368,3.1866817300171415),linewidth(4pt) + dotstyle); label("$M$", (4.094318309882667,3.3701041467208395), NE * labelscalefactor); dot((0.9885564173035511,2.3719712509144015),linewidth(4pt) + dotstyle); label("$E$", (1.0824052888568392,2.550686633647643), NE * labelscalefactor); dot((6.16754816693451,2.207545869029146),linewidth(4pt) + dotstyle); label("$A'$", (6.264667398563043,2.395661698741903), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Let circumcircle of $\triangle BDE$ cut segment $AB$ at a point $E$ and let $A'$ be the reflection of $A$ about $M$. We see that $\overline{ED} || \overline{BC}$ and $\overline{AT} || \overline{BC}$, therefore $A'$ lies on line $\overline{DE}$, with quadrilateral $ADA'T$ being a parallelogram. But, we see that $AE \cdot AB = AD \cdot AC \implies AD = AE$, which means that quadrilateral $AEA'T$ is an isosceles trapezium. Due to converse of Reim's Theorem, we see that $\overline{AT} || \overline{BC} \implies K$ lies on the circumcircle of $\triangle AET$. Therefore, using the fact that $\overline{AD} || \overline{A'T}$, we get that $$\angle CAM = \angle AA'T = \angle AKT$$as desired.

A wise man wrote: I am a simple man. Whenever I see midpoints, I reflect Reflect $A$ over $M$. Let the reflection be $A'$. We see that $ATA'D$ is a parallelogram. We claim that $ATA'K$ is cyclic. Notice that it implies the result. Now we prove our claim. Let $AD$ intersect $AB$ at $X$. It is easy to see that $BCDX$ is cyclic. Therefore $\angle{XKT}=B$. By parallel lines, we also see that $\angle{XAT}=180-B$. Hence, $AXKT$ is cyclic. Now, look at quadrilateral $AXA'T$. Trivially, $AX=AD=A'T$, so $AXA'T$ is a isosceles trapezium, and hence cyclic. Hence, it follows that $AKA'T$ is cyclic.

Let $A'$ be the reflection of $A$ over $M$, thus $ADA'T$ is a parallelogram and let $DA'$ intersect $(BDC)$ at $E$. $AEA'T$ is an isosceles trapezium, hence it's cyclic. $TA' \parallel AD \parallel AC$. $\angle KEA' = \angle KED = \angle KCD = \angle KTA'$ thus $KETA'$ is cyclic. In particular $AEKA'T$ is cyclic.

Prabh2005 wrote: A wise man wrote: I am a simple man. Whenever I see midpoints, I reflect

Let $T'$ and $K'$ be the reflection of $T$ and $K$ over the perpendicular bisector of $BC$, then $K'$ lies on $(BCD)$ since this circle remains unchanged after reflection, $T', A, T$ are collinear, and $T', K', B$ are collinear. By Reim's theorem, since $BCDK'$ is cyclic and $AT' \parallel BC$, then $ADK'T'$ is cyclic. Moreover, $AT = AT'$ and $MT = MD$, so $AM \parallel T'D \implies \angle CAM = ADT'$. Hence, $\angle AKT = \angle AK'T' = \angle ADT' = \angle CAM$, as desired.

Really nice one meh Let $P=(BDC)\cap AB$, $E=AM\cap PD$. First, note that $ADET$ is a parallelogram, because $AT || PD || BC$ (by angle chase) and $M$ is the midpoint of $DT$. Now, note that we just need $ATEK$ to be cyclic. But notice that, because of the triangle being isosceles, $AD=AP=TE$. Thus $ATEP$ is an isosceles trapezoid, so we just need $K\in APET \Leftrightarrow \angle PKT=\angle PET$. To prove this equality let $\angle BAC=2a$, thus $\angle PKT=90-a$ (by cyclic quads) and $\angle PET=\angle DET=90-a$ (by the parallelism). $\blacksquare$

nois 1, btw im back : ɔ

Construct a parallelogram! Let $A'$ be the reflection of $A$ over $M$, and let the circumcircle of $\triangle BCD$ intersect $\overline{AB}$ at $E \neq B$. Since $AE=AD=TA'$, we obtain that $ATA'E$ is an isosceles trapezoid and thus cyclic. Since $\angle KED=\angle KCD=\angle CTA'$, we obtain that $TA'KE$ is cyclic, so $ATA'KE$ is cyclic. Thus, $\angle AKT=\angle AA'T=\angle CAM$.

pad wrote: Here is an unconventional trig approach for the problem. Let $E\not = B$ be the intersection of $(BCD)$ and $AB$. Claim: $A,T,E,K$ concyclic. Proof: Since $AT\parallel BC$, we have \[ \angle ATK=180-\angle TCB=180-\angle KCB=\angle BEK=180-\angle AEK,\]as desired. $\square$ We need to show $\angle DAM=\angle AET$. By the ratio lemma, we have \[ 1=\frac{DM}{MT} =\frac{AD}{AT} \frac{\sin \angle DAM}{\sin \angle MAT} = \frac{AE}{AT} \frac{\sin \angle DAM}{\sin \angle MAT} \implies \frac{AE}{AT} = \frac{\sin \angle MAT}{\sin \angle DAM}. \]But by the Sine Law, \[ \frac{AE}{AT} = \frac{\sin \angle ATE}{\sin \angle AET} \implies \frac{\sin \angle MAT}{\sin \angle DAM} = \frac{\sin \angle ATE}{\sin \angle AET}. \]Here is the key step. Notice that \begin{align*} \angle ATE+\angle AET &= 180-\angle BAT=\angle ABC\\ \angle MAT+\angle DAM &= \angle TAD = \angle TAC=\angle ACB=\angle ABC. \end{align*}So we actually have \[ \frac{\sin (\angle ABC-\angle DAM)}{\angle DAM} = \frac{\sin (\angle ABC-\angle AET)}{\sin \angle AET}.\]But if $\alpha+\beta=C$ for a constant $C \le \pi/2$, then $\tfrac{\sin \alpha}{\sin \beta}$ increases as we increase $\alpha$ since $\sin \alpha$ increases and $\sin \beta$ decreases. Hence $\tfrac{\sin \alpha}{\sin \beta}$ is injective over this range. Therefore, $\angle DAM=\angle AET$, and we are done.

Lol this was a surprise ending. Unique solution!

Let $\odot (BCD)$ meets $AB$ again at $E,$ and $F$ denotes reflection of $A$ over $M,$ so $AEFT$ is an isosceles trapezoid. By Reim theorem $K\in \odot (AET)\implies \angle AKT=\angle AET=\angle AFT=\angle CAM$ $\blacksquare$

moving points because why not