We complex bash. Let $\triangle ABC$ be inscribed in the unit circle, and let line $EF$ intersect the unit circle at $X, Y$. Then $$|a|=|b|=|c|=|x|=|y|=1$$$$a^2 = bc$$$$e = \frac{ac(x+y) - xy(a+c)}{ac-xy}$$$$f = \frac{ab(x+y) - xy(a+b)}{ab-xy}$$Since $AE = BF$, we have $$\frac{a-e}{a-c} = \frac{b-f}{b-a}$$$$\frac{c(a-x)(a-y)}{(a-c)(ac-xy)} = \frac{a(b-x)(b-y)}{(b-a)(ab-xy)}$$Substituting $c = \frac{a^2}b$ gives $$\frac{ab(a-x)(a-y)}{(b-a)(a^3-bxy)} = \frac{a(b-x)(b-y)}{(b-a)(ab-xy)}$$$$b(a-x)(a-y)(ab-xy) = (b-x)(b-y)(a^3-bxy)$$$$a^3b^2 - a^2b^2(x+y) + ab^2xy - a^2bxy + abxy(x+y) - bx^2y^2 = a^3b^2 - a^3b(x+y) + a^3xy - b^3xy + b^2xy(x+y) - bx^2y^2$$$$(a-b)\left[a^2b(x+y) - xy(a+b)^2 + bxy(x+y)\right] = 0$$$$a^2bx + a^2by - a^2xy - 2abxy - b^2xy + bx^2y + bxy^2 = 0$$Let $M$ be the midpoint of one of the arcs $XY$ on the unit circle, so that $$xy = m^2$$$$x + y = \frac{m^2(a+b)^2}{b(a^2+m^2)}$$Then $$e = \frac{ac(x+y) - xy(a+c)}{ac-xy} = \frac{\frac{a^3m^2(a+b)^2}{b^2(a^2+m^2)} - \frac{am^2(a+b)}b}{\frac{a^3}b-m^2} = \frac{am^2(a+b)}{b(a^2+m^2)}$$$$f = \frac{ab(x+y) - xy(a+b)}{ab-xy} = \frac{\frac{am^2(a+b)^2}{a^2+m^2} - m^2(a+b)}{ab-m^2} = \frac{m^2(a+b)}{a^2+m^2}$$$$k = \frac{bc(x+y) - xy(b+c)}{bc-xy} = \frac{\frac{a^2m^2(a+b)^2}{b(a^2+m^2)} - \frac{m^2(a^2+b^2)}b}{a^2-m^2} = \frac{m^2(2a^3b-a^2m^2-b^2m^2)}{b(a^4-m^4)}$$Now point $D$ satisfies $$\frac{d-f}{e-f} = \frac{a-b}{c-b}$$$$\frac{b\left[d(a^2+m^2) - m^2(a+b)\right]}{m^2(a+b)(a-b)} = \frac{b}{a+b}$$$$d(a^2+m^2) - m^2(a+b) = m^2(a-b)$$$$d = \frac{2am^2}{a^2+m^2}$$Then we have the vector $$k' = k-d = -\frac{m^4(a-b)^2}{b(a^4-m^4)}$$And letting $O$ be the circumcenter of $\triangle ABC$, so that $o = 0$, we have $$o' = o-d = -\frac{2am^2}{a^2+m^2}$$Now let $P$ be the projection of $O$ onto line $DK$, and let $p' = p-d$. Then \begin{align*} p' &= \frac{\overline{k'}o' + k'\overline{o'}}{2\overline{k'}} \\ &= \frac{-\frac{2a^3m^2(a-b)^2}{b(a^2+m^2)(a^4-m^4)} + \frac{2am^4(a-b)^2}{b(a^2+m^2)(a^4-m^4)}}{\frac{2a^2(a-b)^2}{b(a^4-m^4)}} \\ &= -\frac{m^2(a^2-m^2)}{a(a^2+m^2)} \end{align*}Then $$p = p'+d = \frac{m^2(a^2+m^2)}{a(a^2+m^2)} = \frac{m^2}a$$So $P$ lies on the unit circle, and thus line $DK$ is tangent to the circumcircle of $\triangle ABC$ at $P$, which is the second intersection of the line through $A$ parallel to line $EF$ with the circumcircle of $\triangle ABC$. $\blacksquare$

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.2) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -8.61, xmax = 14.05, ymin = -6.97, ymax = 7.53; /* image dimensions */ pen zzttqq = rgb(0.6,0.2,0); draw((-2.27,4.47)--(-6.323979716009905,-2.473518449761639)--(1.79,-2.47)--cycle, linewidth(1) + zzttqq); /* draw figures */ draw((-2.27,4.47)--(-6.323979716009905,-2.473518449761639), linewidth(1) + zzttqq); draw((-6.323979716009905,-2.473518449761639)--(1.79,-2.47), linewidth(1) + zzttqq); draw((1.79,-2.47)--(-2.27,4.47), linewidth(1) + zzttqq); draw(circle((-2.267980854884647,-0.18639802173367004), 4.656398459512699), linewidth(1)); draw((-2.27,4.47)--(-3.899510026757201,1.679030698852031), linewidth(1)); draw((-2.27,4.47)--(-6.323979716009905,-2.473518449761639), linewidth(1)); draw((-6.323979716009905,-2.473518449761639)--(1.79,-2.47), linewidth(1)); draw((1.79,-2.47)--(-2.27,4.47), linewidth(1)); draw((0.15807010096592397,0.31955504908780485)--(1.79,-2.47), linewidth(1)); draw((-3.899510026757201,1.679030698852031)--(0.15807010096592397,0.31955504908780485), linewidth(1)); draw((-3.899510026757201,1.679030698852031)--(8.475306577770182,-2.467101063083924), linewidth(1)); draw((8.475306577770182,-2.467101063083924)--(-6.323979716009905,-2.473518449761639), linewidth(1)); draw(circle((-1.4878137916534142,2.1421397292883273), 2.4557582748482045), linewidth(1)); draw((-0.7076467284221815,4.470677480310326)--(8.475306577770182,-2.467101063083924), linewidth(1)); draw((-0.7076467284221815,4.470677480310326)--(-3.899510026757201,1.679030698852031), linewidth(1)); draw((-0.7076467284221815,4.470677480310326)--(0.15807010096592397,0.31955504908780485), linewidth(1)); draw((0.5389335794196923,3.528878276503944)--(-2.267980854884647,-0.18639802173367004), linewidth(1)); draw((-0.7076467284221815,4.470677480310326)--(-2.267980854884647,-0.18639802173367004), linewidth(1)); /* dots and labels */ dot((-2.27,4.47),dotstyle); label("$A$", (-2.19,4.67), NE * labelscalefactor); dot((1.79,-2.47),dotstyle); label("$B$", (1.87,-2.27), NE * labelscalefactor); dot((-6.323979716009905,-2.473518449761639),dotstyle); label("$C$", (-6.25,-2.27), NE * labelscalefactor); dot((-3.899510026757201,1.679030698852031),dotstyle); label("$E$", (-3.81,1.87), NE * labelscalefactor); dot((0.15807010096592397,0.31955504908780485),linewidth(4pt) + dotstyle); label("$F$", (0.23,0.47), NE * labelscalefactor); dot((8.475306577770182,-2.467101063083924),linewidth(4pt) + dotstyle); label("$K$", (8.55,-2.31), NE * labelscalefactor); dot((-0.7076467284221815,4.470677480310326),linewidth(4pt) + dotstyle); label("$D$", (-0.63,4.63), NE * labelscalefactor); dot((0.5389335794196923,3.528878276503944),linewidth(4pt) + dotstyle); label("$G$", (0.61,3.69), NE * labelscalefactor); dot((-2.267980854884647,-0.18639802173367004),linewidth(4pt) + dotstyle); label("$O$", (-2.19,-0.03), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Let $\omega=(ABC)$, $G$ the Miquel point of $CEFB$, and $O$ the center of $\omega$. Note that $D$ lies on $(AEF)$ because $\angle EDF=\angle EAF$. Claim: $O$ lies on $(AEF)$, with $OD$ being a diameter Proof: By PoP on $(ABC)$, and the fact that $\triangle DEF$ is isosceles: \[Pow_{\omega}(E)=AE\cdot EC=BF\cdot AF=Pow_{\omega}(F)\iff\]\[R^2-OE^2=R^2-OF^2\iff\]\[OE=OF\iff\]\[OD\perp EF.\] Let $OD\cap \omega=O'$. Then: \[\angle O'AC=\angle O'AE=\angle O'DE=\angle ODE=\angle A/2=\angle OAC,\]so $O'=O$. As $OD\perp EF$, $OD$ is a diameter $\square$ Claim: $D,G,K$ are collinear Proof: As $GECK$ is cyclic, since $G$ is the Miquel Point, we can say: \[\angle EGK=180-\angle ECK=180-\angle C=180-\angle DFE=180-\angle DGE\]$\square$ As $OD$ is the diameter of $(AEF)$, we can then say $DK\perp OG$, implying the desired result $\blacksquare$

MS_Kekas wrote: Points $E, F$ are selected on sides $AC, AB$ respectively of triangle $ABC$ with $AC=AB$ so that $AE = BF$. Point $D$ is chosen so that $D, A$ are in the same halfplane with respect to line $EF$, and $\triangle DFE \sim \triangle ABC$. Lines $EF, BC$ intersect at point $K$. Prove that the line $DK$ is tangent to the circumscribed circle of $\triangle ABC$. Proposed by Fedir Yudin Let $X$ be the Miquel point of $BCEF$. It suffices to show $DK$ is tangent to $(ABC)$ at $X$. Let $O$ be the circumcentre of $ABC$, then $A, E, O, F, X$ are concyclic by spiral similarity as $\triangle OBF \cong \triangle OAE$. Also $OA=OX$ and $OE=OF$ implies $AX \parallel EF$ and $DA, DX$ symmetric in $OD$. Since $\angle OAD=90^{\circ}$, line $DA$ is tangent to $(ABC)$ so line $DX$ is also a tangent to this circle. Finally, to see line $DK$ is tangent to $(ABC)$ at $X$, it suffices to show that the reflection $L$ of $K$ in line $OD$ lies on line $AD$. This is same as showing the midpoint of $KL$, which coincides with the midpoint of $EF$, lies on the $A$-midline in triangle $ABC$. As this midpoint moves linearly if $E$ moves on $AC$ with constant velocity, and it coincides with midpoints of $AC, AB$ when $E=C, A$ respectively, we are done.