My solution : Let $ T \equiv AD \cap BE $ and $ X \equiv CT \cap AB $ . From $ \angle AED=\angle CFD=180^{\circ}-\angle DGA \Longrightarrow A, D, E, G $ are concyclic , so we get $ \angle BHD=\angle GAD-\angle HDA=\angle GED-\angle GEB=\angle BED $ , hence $ B, D, E, H $ are concyclic $ \Longrightarrow E, F, H $ are collinear (Reim theorem (C-D-B and F-E-H)) . From $ \angle FDE=\angle CAB, \angle FED=\angle CBA \Longrightarrow \triangle FDE \sim \triangle CAB $ , so from Ceva theorem $ \Longrightarrow \frac{BX}{AX}=\frac{BD}{CD} \cdot \frac{CE}{AE}=\frac{GD}{FD} \cdot \frac{FE}{HE}=\frac{FE}{FD}=\frac{CB}{CA} $ , hence $ CX $ is the bisector of $ \angle ACB $ . i.e. the intersection of $ AD $ and $ BE $ lie on the bisector of $ \angle ACB $ Q.E.D

I found this problem in the Menelaus Theorem chapter of Lemmas in Olympiad Geometry, so I was wondering if anyone has a solution using Menelaus Theorem? Thanks!

How to ensure that we can’t have A-B-G ,otherwise GED-GEB=BED won’t work. Thank you!

After finding that $A,E,D,G$ are cyclic, $\frac{\text{sin}(\angle GAQ)}{\text{sin}(\angle EAQ)}=\frac{DG}{DE}=\frac{HE}{DE}$. By trig ceva it suffices to show that $\frac{\text{sin}(\angle DBQ)}{\text{sin}(\angle HBQ)}=\frac{DE}{HE}$. In $\triangle HDE$ $\frac{DE}{HE}=\frac{\text{sin}(\angle DHE)}{\text{sin}(\angle HDE)}$, and we are done because $B,D,E,H$ are cyclic.

Anyone using Menelaus

Has anybody examined the case when $H$ and $G$ are outside the segment $AB$?

Proving the collinearity is trivial. If you add the center of spiral similarity that sends $DG$ to $EH$ (the midarc $ECD$ say $M$) we can prove $MC,DE,AB$ concur by radical at say $S$ and let $AD$ intersect $BE$ at $T$. We have $(S,AT\cap AB;A,B)=-1$ and since $CM$ is exterior bisector $AT$ must be the interior bisector.

Somewhat similar to others, but beatimous. Let $H'=AB\cap FE$. Claim. $H\equiv H'$. Notice that $AEDG$ and $BDEH'$ are cyclic, since $$\measuredangle GAE=\measuredangle BAC=\measuredangle FCE=\measuredangle FDE=\measuredangle GDE$$and $$\measuredangle H'BD=\measuredangle ABC=\measuredangle FCD=\measuredangle FED=\measuredangle H'ED.$$Therefore, we have $$\measuredangle GEB=\measuredangle GED-\measuredangle BED=\measuredangle GAD-\measuredangle BH'D=\measuredangle H'DA,$$and since $\measuredangle HDA = \measuredangle GEB$, we have indeed $H\equiv H'$. $\square$ Let $I$ be the incentre and $Q=AB\cap CI$. Since $AB\parallel CF$, we have $$\frac{HE}{EF}=\frac{AE}{CE}$$and $$\frac{DF}{DG}=\frac{DC}{BD},$$multiplying those together we obtain $$\frac{AC}{BC}=\frac{DF}{EF}=\frac{AE\cdot CD}{CE\cdot BD},$$where the first is true, since $\triangle ABC\sim\triangle DEF$ and we used the fact that $DG=EH$. Therefore we have $$\frac{CD}{DB}\cdot \frac{BQ}{QA}\cdot \frac{AE}{EC}=\frac{CD}{DB}\cdot \frac{BC}{AC}\cdot \frac{AE}{EC}=1$$and we conclude by Ceva's theorem, that $AD$, $BE$ and $CQ$ are concurrent, i.e. $AD$, $BE$ and the angle bisector of $\angle ACB$ are concurrent.

Very nice problem! WLOG, let the diagram be in the figure. (Other cases are similar) Redefine $H$ as the intersection of $EF$ and $AB$. Claim 1: $HEDB$ is cyclic. Proof: Note that $\angle BHE = \angle AHE = \angle EFC = \angle EDC$, hence claim is proven. $\square$ Now we show that this $H$ indeed satisfy $\angle HDA = \angle GEB$. Claim 2: $AEDG$ is cyclic. Proof: Note that $\angle EDF = 180 - \angle ECF = \angle A = \angle GAE$, so claim is proven. $\square$ Claim 3: $\angle BEG = \angle HDA$. Proof: Note that $\angle AEH = \angle FEC = \angle FDC = \angle GDB$. We have $$\angle BEG = 180 - \angle AEG - \angle AEH - \angle BED - \angle DEF = 180 - \angle ADG - \angle GDB - \angle BHD - \angle DBG = 180 - \angle ADG - \angle GDB - \angle HDC = \angle HDA$$. So this claim is also done. $\square$ Now, from $\triangle AEH \sim \triangle CEF$ and $\triangle BDG \sim \triangle CDF$, we have that $HE = \frac{AE \cdot EF}{EC}$ and $DG = \frac{DF \cdot BD}{DC}$. So, $\frac{AE}{EC} \cdot \frac{CD}{DB} = \frac{DF}{EF} = \frac{AC}{BC}$, hence, By Ceva, $Q$ lies on the angle bisector of $\angle ACB$, so we are done. $\square$

Attachments:

Let $X$ denote the intersection of the internal $\angle C-$bisector and side $AB$ and let $N$ denote the major arc midpoint of $DE$ in $(CDE)$. We start off by noticing some cyclic quadrilaterals. [asy][asy] import geometry; size(12cm); pair foot(pair P, pair A, pair B) { return foot(triangle(A,B,P).VC); } path rect = (2,2)--(2,-1.1)--(-3.7,-1.1)--(-3.7,2)--cycle; pair C = dir(140); pair A = dir(220); pair B = dir(320); pair X = intersectionpoint(bisector(line(C,A),line(C,B)),line(A,B)); pair P = (2C+7X)/9; pair D = intersectionpoint(line(A,P),line(B,C)); pair E = intersectionpoint(line(B,P),line(A,C)); pair F = intersectionpoints(parallel(C,line(A,B)),circle(C,D,E))[1]; pair Y = intersectionpoint(line(D,E),line(B,A)); pair G = intersectionpoint(line(F,D),line(A,B)); pair H = intersectionpoint(line(F,E),line(A,B)); pair N = intersectionpoints(line(Y,C),circle(C,D,E))[1]; draw(A--B--C--cycle , red+1.3); draw(C--X,dotted+blue); draw(D--Y,red); draw(A--Y,red); draw(N--Y,red); draw(circumcircle(A,E,D),green); draw(circumcircle(B,E,D),green); draw(circle(F,G,H),purple+dashed); draw(circumcircle(A, B, C) , blue); draw(circumcircle(C, E, D) , orange); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$X$", X, dir(X)); dot("$D$", D, dir(20)); dot("$E$", E, dir(E)); dot("$F$", F, dir(F)); dot("$G$", G, dir(G)); dot("$H$", H, dir(H)); dot("$Y$", Y, dir(Y)); dot("$N$", N, dir(N)); clip(currentpicture,rect); [/asy][/asy] First note that, \[\measuredangle AGF = \measuredangle CFG = \measuredangle CFD = \measuredangle CED\]which implies that $AEDG$ is cyclic. Now, we can deal with the angle condition by showing the following claim. Claim : Quadrilateral $HEDB$ is cyclic, and in turn points $H$ , $E$ and $F$ are collinear. Proof : Note that, \[\measuredangle HDE = \measuredangle ADE + \measuredangle HDA = \measuredangle AGE + \measuredangle GEB = \measuredangle GBE = \measuredangle HBE\]from which it follows that $HEDB$ is cyclic, as claimed. Further, \[\measuredangle FED = \measuredangle FDC = \measuredangle FCB = \measuredangle HBD\]which since $HEDB$ is cyclic, implies that points $H$ , $E$ and $F$ are collinear as desired. Now, in triangles $\triangle NEH$ and $\triangle NDG$, $EH=DG$ and $NE=ND$. Further, \[\measuredangle HEN = \measuredangle FEN = \measuredangle FDN = \measuredangle GDN\]which implies that $\triangle NEH \cong \triangle NDG$. Thus, \[\measuredangle NHF = \measuredangle NHE = \measuredangle NGD = \measuredangle NGF\]from which it follows that $NHGF$ is a cyclic quadrilateral. Further, \[\measuredangle AGN = \measuredangle HGN = \measuredangle HFN = \measuredangle EFN = \measuredangle ECN = \measuredangle ACN\]and thus $CNGA$ must also be cyclic. Now, by the Radical Center Theorem on circles $(CED)$ , $(EDGA)$ and $(CNGA)$, the lines $\overline{DE}$ , $\overline{CN}$ and $\overline{AG}$ intersect at a point $Y$. Further if $Z = \overline{CX} \cap \overline{DE}$, by the Right-angles/Bisectors picture we know \[-1=(YZ;ED)\overset{C}{=}(YX;AB)\]Now, say $P=\overline{AD}\cap \overline{BE}$. Then, letting $X' = \overline{CP} \cap \overline{BC}$, by the Ceva/Menelaus picture we also know, \[-1=(YX';BC)\]which implies that $X'\equiv X$, which implies that indeed segments $AD$ and $BE$ intersect on the internal $\angle ACB-$bisector.

Claim: $HEDB$,$AEDG$ are cyclic, $\implies$ $F,E,H$ are collinear Proof: Let $FE$ intersect extended $BA$ at $H'$. We will now show that $H \equiv H'$. Consider quadrilateral $H'EDB$. $$ \angle H'BD = \angle DCF = \angle DEF $$Hence, $H'EDB$ is cyclic consider quadrilateral $AEDG$ $$\angle AGD = 180^{\circ}-\angle CFD = \angle CED$$Hence $AEDG$ is a cyclic quadrilateral. $$ \begin{aligned} \angle H'DA &= \angle H'DB - \angle BDG -\angle ADG \\ &=\angle H'EB - \angle CDF - \angle AEG \\ &=\angle H'EB - \angle H'EA - \angle AEG \\ &=\angle GEB \\ &=\angle HDA \end{aligned}$$Hence $H\equiv H'$ Let $AD$ and $BE$ intersect at $I$ (This doesnt imply that this is the incenter), $CI$ intersect $AB$ at $J$ . As $CF \parallel HB$, $\triangle CFD \sim \triangle GDB, \triangle CEF \sim \triangle HEA$ $$\therefore \frac{HE}{EF} = \frac{AE}{EC}, \hspace{0.4cm} \frac{BD}{DC} = \frac{GD}{DF}$$ Moreover as $\angle EFD = \angle ECD, \angle CDE= \angle EAB \implies \triangle EFD \sim \triangle ACB $ $$ \frac{EF}{DF} = \frac{BC}{AC}$$From Ceva's Theorem $$ \frac{BJ}{AJ}\cdot\frac{AE}{CE}\cdot\frac{CD}{BD} = 1$$ $$ \frac{BJ}{AJ} = \frac{CE}{AE}\cdot\frac{BD}{CD} = \frac{EF}{HE}\cdot\frac{GD}{DF} = \frac{EF}{DF} = \frac{BC}{AC} $$ Therefore by the Angle Bisector Theorem, the segments $AD$ and $BE$ intersect on the bisector of $\angle ACB$.