Not a nice solution. Notice that $FE$ is perpendicular bisector of $DC$ and therefore $FD=FC$ From the problem condition we have $AF \cdot BC = AC \cdot EC \rightarrow AC \cdot BC = CF \cdot BC + AC \cdot EC$ However, clearly $EC = CF \cdot \cos \angle C$ so plugging this in the equation \[AC \cdot BC = CF \cdot BC + AC \cdot 2 CF \cos \angle C \rightarrow CF = \frac{ AC \cdot BC}{BC + AC \cos \angle C} \] Note that if we prove that $\angle GFE = \angle A$ and $\angle GEF = \angle B$ then we will have that $FE$ is tangent to the circumcircles of $AGF$ and $BGE$ and therefore we will be done. If I prove that $DGFE$ is cylic then we will have $\angle GFE = 180 - \angle GDE = \angle GAC = \angle A$ and $\angle GEF = \angle GDF = 180 - \angle GDB - \angle CDF = 180 - \angle A -\angle C = \angle B$ so this will be enough. We have $\angle DEF = 90$ so proving $\angle DGF = 90$ is enough. Since $\angle GDF = \angle B$ it is enough to show that $\frac{GD}{DF} = \cos \angle B$ Clearly $GD = \frac{AC \cdot BD}{AB}$ and $DF = CF$ so we have that the problem is true iff \[CF \cos \angle B = GD = \frac {AC \cdot BD}{AB}\] \[ \leftrightarrow AB \cdot CF \cos \angle B = AC (BC - CD) =AC( BC - 2CF \cos \angle C)\] Finally, we have that the problem is true iff \[CF = \frac{AC \cdot BC}{AB \cos \angle B + 2 AC \cos \angle C} = \frac{AC \cdot BC}{BC + AC \cos \angle C}\] which is true by the problem condition. The last equality comes from the well known identity $AB \cos \angle B + AC \cos \angle C = BC$

Well, i've got a slightly easier solution... Let $F^{\prime}$ be the reflection of $F$ about the midpoint of $AC$ Since $AF \cdot BC=AC \cdot CE$, then $\frac{F^{\prime}C}{AC}=\frac{CE}{BC} \rightarrow F^{\prime}E \parallel AB$ $\angle BDG=\angle BAC$ ($DGAC$ cyclic quadrilateral) $\angle BAC=\angle EF^{\prime}C$ ($F^{\prime}E \parallel AB$) $\rightarrow \angle GDE= \angle FF^{\prime}F$ (1) $\frac{DE}{F^{\prime}E}=\frac{EC}{F^{\prime}E}=\frac{BC}{AB}$ $\frac{GD}{AC}=\frac{BD}{AB} \rightarrow GD \cdot AB=BD \cdot AC$ And $BD \cdot AC=FF^{\prime} \cdot BC$ $ \iff (BC-2EC)AC=(AC-2AF)BC$ $ \iff AF \cdot BC=AC \cdot CE$ that is the problem condition Then $FF^{\prime} \cdot BC=GD \cdot AB \rightarrow \frac{GD}{FF^{\prime}}=\frac{BC}{AB}=\frac{DE}{F^{\prime}E}$ (2) From (1) and (2) we can deduce that triangles $\triangle GDE$ and $\triangle FF^{\prime}E$ are similar, thus: $\angle DFE=\angle CFE=\angle DGE$ and the quadrilateral $DEFG$ is cyclic, thus: 1) $\angle GFE=\angle BDG=\angle BAC \rightarrow FE$ is tangent to the circumcircle of $\triangle GAF$ 2) $\angle DGE= 90- \angle FCE=90-\angle BGD \rightarrow \angle BGE=90 \rightarrow EF$ is tangent to the circumcircle of $\triangle BGE$ since $BE$ is a diameter and is perpendicular to $EF$

Is there any chance someone could correct the statement i can't figure out were $D$ lies?

paul1703 wrote: Is there any chance someone could correct the statement i can't figure out were $D$ lies? In $]BC[$ But why $\frac{GD}{AC}=\frac{BD}{AB} ?$

Because since $AGDC$ is cyclic me have $\triangle BGD$ similar to $\triangle BCA$

I think it is very easy problem for second question Here is my solution. $\frac{AC}{BC}=\frac{AF}{EC}$. $\frac{AC}{BC}=\frac{sinB}{sinA}$ $\frac{AF}{EC}=\frac{AF}{GF}.\frac{GF}{FD}.\frac{FD}{DE}$ Result of this we'll find that $m(GFE)=A$ easily. In other words $GFDE$ is cyclic quadrilateral. The rest is obvious.

Let $\angle FCD = \angle FDC = x $ and $\angle DGE = y$. We will prove that $x + y = 90$ degrees. Since $AGDC$ is a cyclic quadrilateral, $\angle BGD = x$ and $\angle GAF = \angle GDB$ ... (1). Tangent line of circumcircle of $\triangle AGF $ at point $F$ that meet $BC$ at point $E'$. $\angle GAF = \angle GFE'$ ... (2). From (1) and (2), $\angle GFE' = \angle GDB$. So, $GDFE'$ is cyclic and $\angle FGE' = \angle FDE' = x $ Therefore $ABC \sim FE'G$ (angle-angle-angle similarity) $\frac{AC}{BC}=\frac{GF}{GE'}$ ... (3). Given equality: $AF \cdot BC = AC \cdot EC$ implies $\frac{AC}{BC}=\frac{AF}{DE}$ ... (4) From (3) and (4), we find $\frac{AF}{DE}=\frac{GF}{GE'}$ ... (5) Also $\angle AGF = 180 - (2x + y)$. Let's apply sine theorem in $\triangle AFG$ and cyclic quadrilateral $DGFE'$: $\frac{AF}{FG}=\frac{\sin(2x + y)}{\sin A}$ ... (6) and $\frac{DE}{GE'}=\frac{DF\cdot\cosh}{GE'}=\frac{\sin(x+y)\cdot\cosh}{\sin(GFE')} $ ... (7). By (5), (6) and (7), $\frac{\sin(2x+y)}{ \sin A }=\frac{\sin(x+y)\cosh}{\sin(GFE')}$. Since $\angle GFE' = \angle A$ we yield $\sin(2x + y)= \sin(x + y)\cosh $. Since $\sin((x+y)+x) = \sin(x+y) \cos x + \cos(x+y) \sin x $ we find that $\cos(x+y)\sin x = 0$. So $\cos(x+y) = 0$ and $x + y = 90$. Now we conclude that $E = E'$! Therefore FE line is a tangent of circumcircle of $\triangle BGE$.... $Q.E.D$

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scarface wrote: Tangent line of circumcircle of $\triangle AGD $ at point $F$ that meet $BC$ at point $E'$ I guess you meant $\triangle AGF$

Since $A,G,D,C$ are concyclic, then we have $\triangle ABC \sim \triangle DGB$. $\frac{BC}{AC}=\frac{BG}{DG}$. $\angle GDB = \angle BAC $, $\angle FDC = \angle FCD = \angle BGD \implies \angle GBD = \angle GDF$. We have $AF.BC = AC. EC \implies \frac{BC}{AC}=\frac{CE}{AF} = \frac{BC-CE}{AC-AF}=\frac{BE}{FC}=\frac{BE}{DF}$ Merge the above conclusions $\frac{BC}{AC}=\frac{BG}{DG}=\frac{BE}{DF} \implies \triangle GBE \sim \triangle GAF \implies \angle GEB = \angle GFA$ $\implies G,F,E,D $ are concyclic. So $\angle GFE = \angle GDB = \angle BAC \implies FE $ touches $(GBE)$ and $\angle GBD = \angle GDF = \angle GEF \implies FE $ touches $(AGF)$.