a figure from me with geogebra link

It is sufficient to prove that: $\widehat{EDH}=\widehat{EGD}+\widehat{HFD}$. Now note that $\widehat{HFD}=\widehat{BFD}=\widehat{BED}=\widehat{GED}$, thus $\widehat{EGD}+\widehat{HFD}=\widehat{EGD}+\widehat{DEG}=180-\widehat{EDG}=\widehat{EDH}$ , and we are done.

Let the circle (E, D, G) intersect line BC for the second time at I and A' be the reflection of A across BC. The points E and F are symmetric wrt line BC and with an easy angle-chasing we get that (E, A, D, A') are concyclic, therefore (E, A, D, F) are concyclic. Therefore, $\angle GED= \angle DFH\Rightarrow \angle DFH+\angle DGI= \angle HDI$ and we are done.

We have to prove that $\angle HDE=\angle HFD+\angle DGE$ since $H,D,G$ are collinear this is equivalent with $\angle DEC=\angle AFD$. Let $A'$ be the symmetric of $A$ wrt $BC$ and note $A'\in BE$.We observe that $E,F$ are symmetric wrt $BC$ (this is because if we let $F'$ be the symmetric of $F$ wrt $BC$,since $F=l\cap AB$ we have $F'=l'\cap (AB)'=BE\cap AC=E$). Now we have $\angle AFD=\angle A'ED$ so we have to prove that $(ED$ is the bisector of $A'EC$,but this is easy: note $dist(D,AC)=dist(D,AB)=dist(D,A'E)$ and we are done.

Note that $\angle BEC = \angle BFC =\beta-\gamma$. Moreover, $CD$ is the internal angle bisector of $\angle ACF$ while $BD$ is the external angle bisector of $\angle ABE.$ Hence, $D$ is the incenter in $\triangle ACF$ and the $E-$excenter in $\triangle ABE$. Therefore, $\angle DEG = \angle DFH =\frac{\beta-\gamma}{2}.$ We will be done if we show that $\angle HDE =\angle DGE+\angle HFD.$ However, we know that $\angle HDE = \angle DEG+\angle DGE=\angle HFD+\angle DGE$, so we are done.

Let $A'$ be the reflection of $A$ to $BC$. $A',B,E$ and $A',C,F$ are collinear. \[\frac{EC}{EB}=\frac{sin 180-B}{sin C}=\frac{sin B}{sin C}=\frac{AC}{AB}=\frac{DC}{DB}\]Thus $ED$ is the angle bisector of $\angle CEB$. Similarily $FD$ is the angle bisector of $\angle BFC$. \[\angle FDE=\angle FCE+\angle DFC+\angle CED=\angle B+\angle C\]\[\angle HFD+\angle DGE=\angle B+\angle C-\angle FDH=\angle HDE\]As desired.$\blacksquare$

Quite easy Geo Let $A'$ be the reflection of $A$ about $BC$. Chords $HD$ and $GD$ must inscribe an equal arc in their respective circles. So it is sufficient to show that $\angle HFD=\angle DEG$ or that $\angle AFD=\angle DEC$. But $D$ is the incenter of congruent triangles $EA'C$ and $FAC$ so the result follows.

Another solution along the same lines. Let $A'$ be the reflection of $A$ about $BC$ and let $T$ be the foot of the altitude from $A$. Clearly, $T$ is the midpoint of $AA'$. By Menelaus, $\frac{FB}{AB}\cdot\frac{AT}{TA'}\cdot\frac{CA'}{CF} = 1$. Using that $CA = CA'$, we get $\frac{FB}{FC} = \frac{AB}{AC}$. Therefore $FD$ is the bisector of $\angle BFC$. Analogously, $ED$ is the bisector of $\angle BEC$. By angle chasing, $\angle BEC = \angle BFC = \angle B - \angle C$. Now $\angle DEG = \angle DFH$. Consider the second intersection $X$ of $FD$ with the circumscribed circle of $DEG$. Then, triangles $HDF$ and $GDX$ are similar, so there is a homothety at D between the two circles and we are done.

What do you think about cutoffs?

X.Allaberdiyev wrote: What do you think about cutoffs?

An alternative way to tackle the condition "prove that these circles with a common point are tangent", which does not involve introducing a tangent to any of the circles, is to show (again by angle chase + just that $D$ is incenter of $ACF$) that $D$ is collinear with the centers of the two circles. Btw, despite being an overkill, inverting the diagram at $D$ also works very neatly. Even funnier, we saw a correct solution using moving points!

Asume A^{\prime} is the point such that BC is perpendicular bisector of AA^{\prime}. Then we have \angle CEA^{\prime}=AFA^{\prime}=\angle B-\angle C and \angle EA^{\prime}D=\angle DAC=\frac{1}{2}\angle A, hence EAA^{\prime}F and ADA^{\prime}E are cyclic quadrilaterals. Therefore \angle DFH=\angle DEG.

falantrng wrote: Let $ABC$ be an acute-angled triangle with $AC > AB$ and let $D$ be the foot of the $A$-angle bisector on $BC$. The reflections of lines $AB$ and $AC$ in line $BC$ meet $AC$ and $AB$ at points $E$ and $F$ respectively. A line through $D$ meets $AC$ and $AB$ at $G$ and $H$ respectively such that $G$ lies strictly between $A$ and $C$ while $H$ lies strictly between $B$ and $F$. Prove that the circumcircles of $\triangle EDG$ and $\triangle FDH$ are tangent to each other. Let $d$ is the tangent at $D$ of circle $(DGE)$ Cuz $AD$, $CD$ are internal angle bisectors of $\angle FAC$, $\angle ACF$ respectively, so $D$ is the incenter of $\triangle FAC$ Similar, $D$ is the E-encenter of $\triangle AEB$ Therefore, $ED$, $FD$ in order are internal angle bisectors of $\angle BEC$, $\angle AFC$ In the order hand, since $AB$ and $AC$ in line $BC$ meet $AC$ and $AB$ at points $E$ and $F$ respectively, we see that $\angle BEC = \angle AFC$ Thus, $\angle dDG = \angle DEG = \frac{1}{2} \angle BEC = \frac{1}{2} \angle BFC = \angle DFH$ Lead to $d$ is the tangent at $D$ of circle $(DHF)$ too. Which means $(DEG)$ tangtent $(DFH)$ at $D$

Same solution as above $D$ is clearly the incenter of $\triangle AFC$ and the $E$-excenter of $\triangle AEB$. Moreover, $\triangle BEC$ and $\triangle BFC$ are congruent by symmetry Now let $l$ be the tangent line at $D$ to $(DEG)$, then \[\angle lDH = \angle lDG = \angle DEG = \frac{\angle BEC}{2} = \frac{\angle BFC}{2} = \angle BFD = \angle HFD\]. Therefore, line $l$ is also tangent to $(HFD)$ as desired. $\blacksquare$

Assassino9931 wrote: An alternative way to tackle the condition "prove that these circles with a common point are tangent", which does not involve introducing a tangent to any of the circles, is to show (again by angle chase + just that $D$ is incenter of $ACF$) that $D$ is collinear with the centers of the two circles. Btw, despite being an overkill, inverting the diagram at $D$ also works very neatly. Even funnier, we saw a correct solution using moving points! Can you provide how to use inversion in this figure?

In the language of line arguments, it suffices to show that \[ \measuredangle DE + \measuredangle DG - \measuredangle EG = \measuredangle DF + \measuredangle DH - \measuredangle FH \]since $DG$ and $DH$ are the same line, they cancel and also noting that $E$ and $F$ are reflections across $BC$, so $\measuredangle DF - \measuredangle FH = \measuredangle BE - \measuredangle DE$ so the problem simplifies to proving that \[ \measuredangle DE - \measuredangle AC = \measuredangle BE - \measuredangle DE \]or that $ED$ bisects $\angle AEB$. This follows from observing tht $D$ is the $E$-excenter of $\Delta AEB$.

Let $A'$ be reflection of $A$ in $BC$. We have $$\angle{EAD} = \angle{EAB} + \angle{DAB} = 180^{\circ} - \angle{DAB} = 180^{\circ} - \angle{DA'B}$$Then $E, A, D, A'$ lie on a circle. Similarly, we have $F, A, D, A'$ lie on a circle. So $E, A, D, A', F$ lie on a circle. Suppose that $Dx$ is tangent at $D$ of $(EDG)$. We have $$\angle{FDx} = \angle{EDF} - \angle{EDx} = \angle{EAF} - \angle{EGH} = \angle{AHG} = 180^{\circ} - \angle{FHD}$$Hence $Dx$ tangents $(FDH)$ at $D$

Let the circles $(EDG)$ and $(FDH)$ are $\omega_1$ and $\omega_2$. Let $K$ and $L$ be the second intersection of $BC$ and $\omega_1$, $\omega_2$ respectively. Now by symmetry, $\angle{DEC} = \angle{DFC}$. Now from angle chasing, we get $\angle{DEC} = \angle{DEG} = \angle{DKG}$. Now we know that $DC$ and $DA$ are both angle bisector in $\triangle{AFC}$. So $ FD $ is the bisector of $\angle{AFC}$. Hence $\angle{DFC} = \angle{DFB}$. Now if we do some angle chasing, we get $\angle{DFB} = \angle{DFH} = \angle{DLH}$. And by symmetry, $\angle{DEC} = \angle{DFC}$. Finally, we conclude that by all of these, $\angle{DKG} = \angle{DLH}$ and that shows $KG$ $||$ $LH$. By this condition, we are going to prove that $\omega_1$ and $\omega_2$ are homothetic centered $D$ and this ends the proof. Assume not, so these circles aren't tangent to each other and they aren't homothetic centered at some point which is on both $\omega_1$ and $\omega_2$. Now, we take $M$ as the second intersection of $\omega_1$ and $\omega_2$. (You don't have to show this point on the shape, because you can't.) İf we take the spiral homothety such that $\omega_1 \rightarrow \omega_2$ which is centered at $M$, surely $K \rightarrow L$ and $G \rightarrow H$. Hence $KG \rightarrow LH$. So this spiral homothety has $180$ degree angle. That means it is actually a homothety which has a negative ratio. And that contradicts our assumption. $\blacksquare$

let $ER$ intersect $CF$ at $A'$, it's known that $A'$ is the reflection of $A$ about $BC$, and since $\measuredangle EA'C=\measuredangle \frac{1}{2 } ABC $ and $\measuredangle ECA$ =$\frac{1}{2} \measuredangle ACB$ thus we have that $D$ is the incenter of $\triangle EA'C$ thus $\measuredangle CEB=\measuredangle CFB$ by doing the same thing to $\triangle CBF$ we get that $\widehat{GD}=\widehat{DH}$ because it's known that $\measuredangle DFH=\measuredangle GFD$ because $\triangle CBF \equiv \triangle BCF $ and thus we are done $\quad \square $ [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(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -52.84492215023231, xmax = 50.96732711565892, ymin = -28.95316569900705, ymax = 20.59704405308735; /* image dimensions */ draw(arc((-20.05320608200209,-27.82763712835592),4.939821524365248,59.88882786987366,76.25774675333359)--(-20.05320608200209,-27.82763712835592)--cycle, linewidth(0.8)); draw(arc((-18.974370747974863,15.785898800233332),4.939821524365248,-62.722815409602404,-46.35389652614247)--(-18.974370747974863,15.785898800233332)--cycle, linewidth(0.8)); /* draw figures */ draw((-11.343744198496259,7.785851413180044)--(-14.749072327731113,-6.138730373020776), linewidth(0.4)); draw((-14.749072327731113,-6.138730373020776)--(2.341062129234857,-6.561476213799192), linewidth(0.4)); draw((2.341062129234857,-6.561476213799192)--(-11.343744198496259,7.785851413180044), linewidth(0.4)); draw((-11.343744198496259,7.785851413180044)--(-7.577811376086417,-6.3161204777082665), linewidth(0.4)); draw((-12.036371586864286,-20.214649251689206)--(-14.749072327731113,-6.138730373020777), linewidth(0.4)); draw(circle((-13.120272392807134,4.815234612631612), 12.434867923267397), linewidth(0.4)); draw(circle((2.48440758612562,-26.52485661987126), 22.575235699922708), linewidth(0.4)); draw((-20.05320608200209,-27.82763712835592)--(-12.036371586864286,-20.214649251689206), linewidth(0.4)); draw((-12.036371586864286,-20.214649251689206)--(2.341062129234857,-6.561476213799192), linewidth(0.4)); draw((-16.9082686801178,-14.967807942127294)--(-7.577811376086417,-6.3161204777082665), linewidth(0.4)); draw((-7.577811376086417,-6.3161204777082665)--(-2.4384185572680024,-1.5506069582278386), linewidth(0.4)); draw((-14.749072327731113,-6.138730373020776)--(-18.974370747974863,15.785898800233332), linewidth(0.4)); draw((-11.343744198496259,7.785851413180044)--(-18.974370747974863,15.785898800233332), linewidth(0.4)); draw((-14.749072327731113,-6.138730373020776)--(-20.05320608200209,-27.82763712835592), linewidth(0.4)); draw((-7.577811376086417,-6.3161204777082665)--(-18.974370747974863,15.785898800233332), linewidth(0.4)); draw((-7.577811376086417,-6.3161204777082665)--(-20.05320608200209,-27.82763712835592), linewidth(0.4)); /* dots and labels */ dot((-11.343744198496259,7.785851413180044),linewidth(1pt) + dotstyle); label("$A$", (-11.04643232868021,7.905502598179735), NE * labelscalefactor); dot((-7.577811376086417,-6.3161204777082665),linewidth(1pt) + dotstyle); label("$D$", (-7.322566871851023,-7.521940008684013), NE * labelscalefactor); dot((-18.974370747974863,15.785898800233332),linewidth(1pt) + dotstyle); label("$E$", (-18.646157750780592,15.961211545606126), NE * labelscalefactor); dot((-20.05320608200209,-27.82763712835592),linewidth(1pt) + dotstyle); label("$F$", (-19.558124801432637,-28.345187665239024), NE * labelscalefactor); dot((-2.4384185572680024,-1.5506069582278386),linewidth(1pt) + dotstyle); label("$G$", (-2.154753584822762,-1.3661624167827144), NE * labelscalefactor); dot((-16.9082686801178,-14.967807942127294),linewidth(1pt) + dotstyle); label("$H$", (-18.114176971233565,-14.66568190545836), NE * labelscalefactor); dot((-12.036371586864288,-20.214649251689206),linewidth(1pt) + dotstyle); label("$A'$", (-11.730407616669245,-21.04945126002267), NE * labelscalefactor); dot((2.3410621292348566,-6.5614762137991915),linewidth(1pt) + dotstyle); label("$C$", (2.633073431100479,-6.381981195368958), NE * labelscalefactor); dot((-14.749072327731112,-6.138730373020776),linewidth(1pt) + dotstyle); label("$B$", (-16.21424561570847,-6.22998668692695), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy]

From angle bisector theorem applies a couple times we have: \[ \frac{BF}{FC}=\frac{BA}{AC}=\frac{BA'}{A'C}=\frac{BE}{EC} \implies FA'DAE \; \text{cyclic} \]Because it's just the $A$-apollonian circle, to finish just note that $\angle AFD=\angle AED$ thus we are done .

Just angle chasing