Nice and easy. Let $X$ be the intersection of $BE$ and $CF$. Notice that $E$ and $F$ lie on the circle with diameter $XD$. We will prove that $A$ also lies on the aforementioned circle which clearly suffices. Notice that $$\angle BXC = \angle EBF = 180^{\circ} - \angle EDF = 180^{\circ} - \angle EDA - \angle ADF = 180^{\circ} - \angle CBA - \angle BCA = \angle BAC$$using the two tangencies, meaning that $X$ lies on the circumcircle of $\triangle ABC$. Then $$\angle XAD = \angle XAB + \angle BAD = XCB + \angle FDC = \angle FCD + \angle FDC = 90^{\circ}$$meaning that $A$ also lies on the circle with diameter $XD$, as desired. $\blacksquare$

Solved with MrOreoJuice, SatisfiedMagma and Fakesolver19. Let $X= BE \cap CF$ Claim: Points $E$, $X$, $F$ and $D$ are concyclic. $\measuredangle DEA= 90^{\circ}=\measuredangle AFD$. Claim: Points $A$, $X$, $C$ and $B$ are concyclic. $\measuredangle EXF = \measuredangle BAC$. Claim: Points $A$, $X$, $D$ and $E$ are concyclic. $\measuredangle ADE=\measuredangle ACB=\measuredangle AXE$. The above claims imply that points $A$, $D$, $E$, $F$ and $X$ are concyclic, so we are done.

[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 = -16.357415640299983, xmax = 22.327334547161975, ymin = -12.89121383076778, ymax = 10.238832625965346; /* image dimensions */ pen qqwuqq = rgb(0,0.39215686274509803,0); pen ffwwqq = rgb(1,0.4,0); /* dots and labels */ dot((1,6),dotstyle); label("$A$", (1.0911237719299203,6.249152123603082), NE * labelscalefactor); dot((-6.619492600422831,-2.1819450317124702),dotstyle); label("$B$", (-6.509470096494394,-1.9322180204815604), NE * labelscalefactor); dot((8.06,-5.38),dotstyle); label("$C$", (8.161443649533933,-5.1391131078233805), NE * labelscalefactor); dot((-1.8594274338429146,-3.218966573922643),dotstyle); label("$D$", (-1.7622553215316994,-2.967514859702148), NE * labelscalefactor); dot((-3.391001794903157,-0.4171408524866453),linewidth(4pt) + dotstyle); label("$E$", (-3.302575009152574,-0.21514033592058612), NE * labelscalefactor); dot((7.0357301393792495,-1.0934642329194084),linewidth(4pt) + dotstyle); label("$F$", (7.126146810313346,-0.896921181260973), NE * labelscalefactor); dot((5.688208854811421,4.545868139607652),linewidth(4pt) + dotstyle); label("$T$", (5.787836262052586,4.7593347208222365), NE * labelscalefactor); /* draw figures */ draw((1,6)--(-6.619492600422831,-2.1819450317124702), linewidth(0.8)); draw((-6.619492600422831,-2.1819450317124702)--(8.06,-5.38), linewidth(0.8)); draw((8.06,-5.38)--(1,6), linewidth(0.8)); draw(circle((4.038006010778352,0.004773500535603293), 6.72102829196586), linewidth(0.8) + linetype("4 4") + qqwuqq); draw(circle((-4.239460017132872,-2.7004558028175567), 2.4358588869788536), linewidth(0.8) + linetype("2 2") + ffwwqq); draw((-6.619492600422831,-2.1819450317124702)--(-3.391001794903157,-0.4171408524866453), linewidth(0.8)); draw((-3.391001794903157,-0.4171408524866453)--(-1.8594274338429146,-3.218966573922643), linewidth(0.8)); draw((1,6)--(-1.8594274338429146,-3.218966573922643), linewidth(0.8)); draw(circle((-3.1634780906986264,2.238442647617127), 5.611048344738542), linewidth(0.8) + linetype("4 4") + qqwuqq); draw(circle((1.91439071048425,0.6634507828425029), 5.4143206331509575), linewidth(0.8) + blue); draw(circle((3.100286283078543,-4.299483286961322), 5.076049302457057), linewidth(0.8) + linetype("2 2") + ffwwqq); draw((-1.8594274338429146,-3.218966573922643)--(7.0357301393792495,-1.0934642329194084), linewidth(0.8)); draw((7.0357301393792495,-1.0934642329194084)--(1,6), linewidth(0.8)); draw((1,6)--(-3.391001794903157,-0.4171408524866453), linewidth(0.8)); draw((-3.391001794903157,-0.4171408524866453)--(7.0357301393792495,-1.0934642329194084), linewidth(0.8)); draw((7.0357301393792495,-1.0934642329194084)--(8.06,-5.38), linewidth(0.8)); draw(circle((1.1554093314987617,-1.7835509771193976), 7.785102303356918), linewidth(0.8) + linetype("4 4") + red); draw((-3.391001794903157,-0.4171408524866453)--(5.688208854811421,4.545868139607652), linewidth(0.8)); draw((5.688208854811421,4.545868139607652)--(7.0357301393792495,-1.0934642329194084), linewidth(0.8)); draw((1,6)--(5.688208854811421,4.545868139607652), linewidth(0.8)); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Nice Assume that $F$ is outside $ABC$. Let $T$ be the intersection of $BE$ and $CF$,and let $\alpha , \beta ,\gamma , x$ be the angles $\angle A, \angle B, \angle C, \angle FDC$. Notice that $\angle BCA =\angle DCA = \angle EDA$ and that $\angle CBA = \angle DBA = \angle FDA$ by the conditions of our problem, then this implies that $\angle EDF = 180-\alpha$. Since $DCF$ is a right-angled triangle then we have that $\angle DCF=90-x$, also notice that $\angle DBE = 90-\angle EDB = 90+x-\alpha$. This implies that $\angle BTC = 180-(\angle TBC + \angle TCA)=180-(\angle DBE + \angle DCF)=\alpha$, thus umplying that $T$ is on the circumcircle of $ABC$ and the circumcircle of $EDF$. Now notice that $\angle ATE = \angle ATB =\angle ACB =\angle DCA = \angle EDA$, implying that $T$ is on the circumcircle of $AED$. Similarly we have that $T$ is on the circumcircle of $AFD$, which combined with that $T$ is on the circumcircle of $EDF$, gives us that $TAEDF$ is a cyclic pentagon, which implies that $AEDF$ is a cylic quadrilateral.

Let $\angle CAD=\alpha$ and $\angle BAD=\beta$ $\implies$ $\angle EDB=\alpha$ and $\angle FDC=\beta$. It is well-known that $\frac{BD}{DC}=\frac{AB\cdot sin(\beta)}{AC\cdot sin(\alpha)}$ $\implies$ $\frac{AB}{AC}=\frac{BD\cdot sin(\alpha)}{DC\cdot sin(\beta)}=\frac{BE}{CF}$ $\cdots$ $(\star)$ If $\angle ACD=\gamma$ then $\angle ABE=\angle ACF=90-\beta-\gamma$ $\cdots$ $(\star$ $\star)$ From $(\star)$ and $(\star$ $\star)$ we get $\triangle ABE \sim \triangle ACF$ $\implies$ $E$ and $F$ are isogonal wrt $\angle ABC$ $\implies$ $\angle EAF=\angle BAC=180-\angle EDF$ $\implies$ $AEDF$ is cyclic as desired

Let $BE \cap CF = X$. Trivially $EDFX$ Is cyclic. Next, let $\angle{ABD}=\angle{ADF}=\alpha, \angle{ACD}=\angle{ADE}=\beta$. Note that $\angle{EXF}=180-\alpha-\beta=A \implies ABCX$ is cyclic. Therefore, $\angle{AXB}=\beta=\angle{AXE}=\angle{ADE} \implies XAED$ is cyclic. Now $\angle{XAD}=90 ,\angle{XFD}=90$. Therefore, $XAFD$ is cyclic. This implies $X,A,E,F,D$ concylic, as desired.

As an alternative short solution, one can invert w.r.t. $D$. All the circles disappear and we only need to prove that $A',E',F'$ are collinear which is almost trivial.

Very easy problem by Ptolemy sinus lemma.it suffices to prove the following. $$DE \cdot sin \angle ABC + DF \cdot sin \angle ACB = DA \cdot \angle BAC$$

Trivial by realizing, wrt <ABC, E and F are isogonal