$MG \times ME = MB \times MC = (AB-AM) \times (AC-AM)$ $MG \times ME = AB \times AC - AB \times AM - AC \times AM + AM^2$ Now if we show that $MG \times MF = AM^2$ then the problem is finished, and it is true since the triangles $AMG$ and $AME$ are similar, thus $\frac{MG}{AM} = \frac{AM}{ME}$ equivalent to $AM^2 = MG \times MF$. So we have desired $AB \times AC= AB \times AM + AC \times AM$ which is $\frac{1}{AM} = \frac{1}{AB} + \frac{1}{AC}$. the problem is finished.

Since the lines $DF$ and $AC$ are parallel, it follows that $\angle DFA=\angle CAF.$ On the other hand, $\angle DFA=\angle DEG,$ because both angles subtend the arc $DG.$ We deduce that $\angle CAF=\angle DEG,$ hence triangles $AMG$ and $EMA$ are similar. This implies \[ \frac{AM}{MG}=\frac{EM}{AM}, \] that is, $AM^{2}=MG\cdot ME.$ From the power of a point theorem we have \[ MG\cdot ME=MB\cdot MC, \] hence \[ AM^{2}=MB\cdot MC. \] Writing the latter as \[ AM^{2}=\left( AB-AM\right) \left( AC-AM\right) =AB\cdot AC-AM\left( AB+AC\right) +AM^{2}, \] gives \[ AM\left( AB+AC\right) =AB\cdot AC, \] or, equivalently, \[ \frac{1}{AM}=\frac{1}{AB}+\frac{1}{AC}, \] as desired.

Proof. $DF\parallel AC$ $\Longleftrightarrow$ $(\widehat {AGM}\equiv\widehat {EGF}\equiv\widehat {EDF}\equiv\widehat {EAM})$ $\Longleftrightarrow$ $\widehat {AGM}\equiv\widehat {EAM}$ $\Longleftrightarrow$ $\triangle AGM\sim\triangle EAM$ $\Longleftrightarrow$ $MA^2=ME\cdot MG$ $\Longleftrightarrow$ $MA^2=MB\cdot MC$ $\Longleftrightarrow$ $\boxed {\ \frac{1}{AM}=\frac{1}{AB}+\frac{1}{AC}\ }\ .$ Remark 1. If the point $N$ is the reflection of the point $A$ w.r.t. the point $M$ then $\frac{1}{AM}=\frac{1}{AB}+\frac{1}{AC}$ $\Longleftrightarrow$ $\frac{2}{AN}=\frac{1}{AB}+\frac{1}{AC}$ $\Longleftrightarrow$ the division $(A,B,N,C)$ is harmonically, i.e. $\frac{NB}{NC}=\frac{AB}{AC}\ .$ Remark 2. See the following proposed problems: nr. 1.9., the page 9., about the characterization of a harmonical division; nr. 5.103., the page 58., with three extensions in the my book Geometrie plana (sintetica, vectoriala, analitica). Culegere de probleme., Editura GIL, Zalau, 2002, Romania.

Consider the inversion of pole $A$ wich invariates the circle. If $M'$ is the inverse of $M$ then $M'FDA$ is an isosceles trapezoid. Thus $AM'=AB+AC$ and the conclusion follows.

easy, right ?

A generalization. Consider two circles $C_k=C(O_k)\ ,\ k\in \overline {1,2}\ ,$ a fixed point $A\in C_1$ and a mobile point $M\in C_2\ .$ Denote: the tangent $d\equiv MM$ in the point $M$ to the circle $C_ 2\ ;$ the second intersection $N$ of the line $AM$ with the circle $C_1\ ;$ the point $P$ of the circle $C_1$ for which $NP\parallel d\ ;$ the second intersection $R$ of the line $MP$ with the circle $C_1\ ;$ the intersection $L\in AR\cap d\ .$ Prove that the relation $LA\cdot LR=LM^2$ and ascertain the geometrical locus of the point $L$ when the point $M$ covers the circle $C_2\ .$

This probelm have a simple projective solution. From Desargue th for the cyclic quadrilateral $DEFG$ and the line $AM$ we have that this line cuts the pairs of opposite sides, the diagonals and the circumcircle of $DEFG$ in point in the same involution. Denote the ecuation of this involution with $axx'+b(x+x')+c=0$ with origin in $A$. Since $(A,A)$ is a pair of points in involution we have that, $c=0$, hence the involution is $\frac 1 x +\frac 1 {x'}=-\frac a b$ which is const. Hence, $-\frac a b=\frac 1 {AM}=\frac 1 {AM}+\frac 1 {A \infty}=\frac 1 {AB}+\frac 1 {BC}$ q.e.d.

We consider the following: Perspectivity from E gives $(A,B;M,C)=(D,B;G,C)$. Perspectivity from F gives $(D,B;G,C)=(A,C;B,\infty)$ Now, compare the cross ratios so obtained. We get the conclusion.

Let $P$ be the reflection of $A$ in $M$ and let $N$ be the midpoint of $\overline{BC}.$ Since $CFDB$ is an isoceles trapezoid, we have $\widehat{CF} = \widehat{DB}.$ Therefore, \[\measuredangle MAG = \frac{1}{2}\left(\widehat{CF} - \widehat{GB}\right) = \frac{1}{2}\left(\widehat{DB} - \widehat{GB}\right) = \frac{1}{2}\widehat{DG} = \measuredangle DEG = \measuredangle AEG.\]Thus, by the Alternate Segment Theorem, $MA$ is tangent to $\odot (AGE)$, and in particular, $MA^2 = MG \cdot ME = MB \cdot MC.$ Therefore, it is well-known (Lemma 1.5) that $(B, C; A, P)$ is harmonic. By the same Lemma 1.5, we have $AB \cdot AC = AP \cdot AN = 2 \cdot AM \cdot AN.$ Hence, \[\frac{1}{AM} = \frac{2 \cdot AN}{AB \cdot AC} = \frac{AB + AC}{AB \cdot AC} = \frac{1}{AB} + \frac{1}{AC}. \; \square\]

Dear MLs, M.T.