Nice and easy. Let $O$ be the center of the circumcircle of $\triangle ABC$ and also the midpoint of $AD$. Notice that if we let $DF$ intersect the circumcircle again at a point $C'$, then $$\angle C'DA = \angle FDA = \angle DAE = \angle DAC$$which ultimately means that $C'$ is the $C$ antipode, similarly, $DE$ intersects the circumcircle at the $B$ antipode, say $B'$. Moreover, notice that as $AFDE$ is a parallelogram that $O$ lies on $EF$. Then by Pascal's Theorem on $DDABCC'$, we have that $G'$ which is the intersection of $D$ tangent and $BC$ also lies on $FO$ which coincides with $EF$ and therefore $G = G'$. We therefore get that $GD$ is tangent to the circumcircle and therefore $$GD \perp AD$$because $AD$ is a diameter, as desired. $\blacksquare$

Let $DF$ and $DE$ meet $(ABC)$ at $Y$ and $X$. Pretty obvious angle chasing gives $CY$ diameter, so Pascal on $DDYCBA$ gives $F$, $O$, $G$ colinear ($O$ is center of $ABC$), and analogously for the other three points and this finishes. Comment: The thought of Pascal already solves this.

Clearly, $\angle BDE = \angle CDF = 90^{\circ}$. Applying DDIT on point $D$ and quadrilateral $BFCE$, we get an involution swapping $(DB, DE)$, $(DC, DF)$ and $(DA,DG)$. But rotating by $90^{\circ}$ is an involution, so we are done!

Isn't this well known? Let $O$ be the circumcenter, $DE \cap (ABC) = B'$ which is the $B$ antipode and $DF \cap (ABC) = C'$ which is the $C$ antipode. Redefine $G$ to be the intersection of tangent at $D$ to $(ABC)$ with $BC$. Pascal on $ABB'DC'C$ gives $\overline{F-O-E}$ and Pascal on $DDC'CBA$ gives $\overline{G-F-O}$ which finishes.

@above their solutions are very interesting. I have a different solution: Inversion on $D$ with reflexion on the angle bisector and ratio $\sqrt{DB \times DC}$. Obviously $F, O, E$ are collinears, but we can prove that $O, E, G$ are collinears. Then it's sufficient to solve the following problem: Let $G'$ in the circumcircle of $\triangle BDC$ such that $DG$ is parallel to $BC$. $O'$ is the reflexion of $D$ on $BC$ and $A'$ is the projection of $D$ to $BC$. Let $E'$ the intersection of $(O'DG')$ and $(BA'D)$. Prove that $\angle E'DC=90^o$. For that it's sufficient to prove that $M$ (the midpoint of $BC$) is the center of $(O'E'DG')$.

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(7cm); 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 = -1.9672456517856833, xmax = 2.5083672179472067, ymin = -1.3320184173813823, ymax = 1.2698195812905677; /* image dimensions */ draw((-0.32828582969112824,0.9445784319070638)--(-0.8,-0.6)--(0.8,-0.6)--cycle, linewidth(0.8)); /* draw figures */ draw(circle((0,0), 1), linewidth(0.8)); draw((-0.32828582969112824,0.9445784319070638)--(-0.8,-0.6), linewidth(0.8)); draw((0.8,-0.6)--(-0.32828582969112824,0.9445784319070638), linewidth(0.8)); draw((0.32828582969112824,-0.9445784319070638)--(0.8,0.6), linewidth(0.8)); draw((0.32828582969112824,-0.9445784319070638)--(-0.8,0.6), linewidth(0.8)); draw((-0.8,-0.6)--(1.319743046062613,-0.6), linewidth(0.8)); draw((-0.5415661760697021,0.2462143722683461)--(1.319743046062613,-0.6), linewidth(0.8) + linetype("4 4")); draw((0.32828582969112824,-0.9445784319070638)--(1.319743046062613,-0.6), linewidth(0.8)); /* dots and labels */ dot((0,0),dotstyle); label("$O$", (-0.05,-0.16), NE * labelscalefactor); dot((-0.32828582969112824,0.9445784319070638),dotstyle); label("$A$", (-0.40483917928935,0.9735148200535235), NE * labelscalefactor); dot((-0.8,-0.6),dotstyle); label("$B$", (-0.9638631599410679,-0.6745040424458462), NE * labelscalefactor); dot((0.8,-0.6),dotstyle); label("$C$", (0.8010874032004235,-0.742873924), NE * labelscalefactor); dot((0.32828582969112824,-0.9445784319070638),dotstyle); label("$D$", (0.3363983656,-1.1039056092407), NE * labelscalefactor); dot((0.8,0.6),dotstyle); label("$P$", (0.8123752036285014,0.6292369069971483), NE * labelscalefactor); dot((-0.8,0.6),dotstyle); label("$Q$", (-0.9,0.6320588571041678), NE * labelscalefactor); dot((0.5415661760697021,-0.2462143722683461),linewidth(4pt) + dotstyle); label("$E$", (0.38,-0.38), NE * labelscalefactor); dot((-0.5415661760697021,0.2462143722683461),linewidth(4pt) + dotstyle); label("$F$", (-0.70657002022954,0.211588291158267), NE * labelscalefactor); dot((1.319743046062613,-0.6),linewidth(4pt) + dotstyle); label("$G'$", (1.342901823748163,-0.6688601422318072), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Let $O$ be the centre of $\odot (ABC)$. Let $P, Q$ respectively be the reflections of $B, C$ over $O$. Then $AB\parallel DP$ and $AC \parallel DQ$. Notice that $O$ is the midpoint of $\overline{EF}$ since $AFDE$ is a parallelogram. Suppose the tangent to $\odot(ABC)$ at $D$ meet $BC$ at $G'$. Then $G'\in OF = EF$ by Pascal's theorem on $ABCQDD$. Thus $G'=G$, and we are done. $\blacksquare$

Menelaus theorem

some syntetic and easy complex bash Note $O$ be circumcenter.Let $X$ be second intersection of $DE$ with $(ABC)$. Just easy angle chasing yields $BX$ is diameter Now Let $(ABC)$ be unit circle $|a|=|b|=|c|=1$. Then $d=-a , x=-b$ $e=\frac{dx(a+c)-ac(d+x)}{dx-ac}=\frac{ab+2bc+ac}{b-c}$ also notice that $\bar{e}=\frac{b+2a+c}{a(c-b)}$ Since $G$ is intersection of $EO$ with $BC$. $g=\frac{e(\bar{c}b-\bar{b}c)}{\bar{e}(b-c)-e(\bar{b}-\bar{c})}=\frac{a(ab+bc+2bc)}{a^2-bc}$ and notice $\bar{g}=\frac{2a+b+c}{bc-a^2}$ Thus we just need to show that $\frac{a-d}{g-d}+\frac{\bar{a}-\bar{d}}{\bar{g}-\bar{d}}=0 \iff \frac{2a}{\frac{a(ab+2bc+ac)}{a^2-bc}+a}+\frac{\frac{2}{a}}{\frac{2a+b+c}{bc-a^2}+\frac{1}{a}}=0 \iff \frac{2(a^2-bc)}{(a+b)(a+c)}+\frac{2(bc-a^2)}{(a+b)(a+c)}=0 \iff 0=0$ so we are done

Let $O$ be the circumcenter of $(ABC)$ and $T$ be a point on $(ABC)$ such that $AT \parallel EF$. It is obvious that $AFDE$ is a parallelogram and $O$ is the midpoint of $EF$. In the trapezoid $ATEF$ we have that $O$ is the midpoint of $EF$ and $OA=OT$, hence it is isosceles trapezoid and hence $ATEF$ is cyclic. From the cyclic quadrilaterals we get that: $\angle TCG = \angle BAT = \angle FAT = \angle TEG$ and hence $TECG$ is alco cyclic. So we get that: $\angle CTG = \angle CEG = \angle FEA = \angle EAT = \angle CAT$, from where we conclude that $GT$ is tangent to $ABC$. Also: $AT \perp TD => EF \perp TD$ as $EF \parallel AT$. Now, using that $OT=OD$ and $O$ lies on $EF$, we get that $EF$ is the perpendicular bisector of $DT$. Hence, $GD=GT$ and so $GD$ is also tangent to $(ABC)$, i.e. $AD \perp DG$, as needed.

Attachments:

bluds problem is in problem solving tactics literally orthotransversal (xooks)

pascal on $ACYDXB$ and $DDXBCA$, $DE\cap (ABC)=X, DF\cap (ABC)=Y$.

Let $O$ be the circumcenter of $\Delta ABC$. Let $G'$ be the intersection of line $BC$ and the tangent to $(ABC)$ at $D$. Let $B'$ and $C'$ be the $B$- and $C$- antipode in $(ABC)$ respectively. Note that $D-E-B'$ collinear and $D-F-C'$ collinear. By Pascal on $DDB'BCA$, $O-E-G'$ collinear, and by Pascal on $DDC'CBA$, $O-F-G'$ collinear. It follows that $F-E-G'$ collinear, so $G=G'$, so $AD \perp DG$.