19-th Macedonian Mathematical Olympiad

Let $ABCD$ be such a quadrangle. Notice that, according to the conditions of the exercise, the line $EG$ intersects the side $BC$ in a point inside of $BC$. Let us denote that point by $H$. We will consider two cases:

Case 1: The lines $AB$ and $CD$ are not parallel.

Let them intersect at point $Q$. By Menelaus's theorem for the triangle $EQG$ and the line $CB$ and $DA$ we have $$\frac{\overline{QC}}{\overline{CG}}\cdot \frac{\overline{GH}}{\overline{HE}}\cdot \frac{\overline{EB}}{\overline{BQ}}=1,\frac{\overline{QD}}{\overline{DG}}\cdot \frac{\overline{GF}}{\overline{FE}}\cdot \frac{\overline{EA}}{\overline{AQ}}=1.$$ If we multiply the last two equalities and use that $\overline{QC}\cdot \overline{QD}=\overline{BQ}\cdot \overline{AQ}$, since that is the degree of point $Q$ with respect to circle $k$, we get $$\frac{\overline{GH}}{\overline{CG}}\cdot \frac{\overline{EB}}{\overline{HE}}\cdot \frac{\overline{GF}}{\overline{DG}}\cdot \frac{\overline{EA}}{\overline{FE}}=1,$$ or $$\frac{\overline{GH}}{\overline{HE}}=\frac{\overline{CG}\cdot \overline{DG}}{\overline{EB}\cdot \overline{EA}}\cdot \frac{\overline{FE}}{\overline{GF}}(1)$$ Notice that $\overline{CG}\cdot \overline{DG}$ and $\overline{EB}\cdot \overline{EA}$ are the degrees of points $G$ and $E$ with respect to circle $k$ respectively and that they do not depend on the choice of the quadrangle $ABCD$. It is clear that $\overline{FE}$ and $\overline{GF}$ do not depend on the choice of the quadrangle $ABCD$.

Case 2: The lines $AB$ and $CD$ are parallel. It is clear that $△GCH\sim △EBH$ and $△GFD\sim △EFA$. From the similarity, we have $$\frac{\overline{GH}}{\overline{EH}}=\frac{\overline{GC}}{\overline{EB}},\frac{\overline{EA}}{\overline{FE}}=\frac{\overline{GD}}{\overline{GF}}$$ If we multiply the last two equalities, we get $$\frac{\overline{GH}}{\overline{EH}}\cdot \frac{\overline{EA}}{\overline{FE}}=\frac{\overline{GC}}{\overline{EB}}\cdot \frac{\overline{GD}}{\overline{GF}},$$ or $$\frac{\overline{GH}}{\overline{EH}}=\frac{\overline{GC}\cdot \overline{GD}}{\overline{EB}\cdot \overline{EA}}\cdot \frac{\overline{FE}}{\overline{GF}}(2)$$ which is actually the same as in the first case. We conclude that the side $BC$ intersects the line $EG$ in a point $H$ for which (1) (which is the same as (2)) holds. Since $E$ and $G$ lie outside the circle and $F$ lies inside the circle and the extensions of sides $AB$, $AD$ and $DC$ pass through $E$, $F$ and $G$ respectively, if $ABCD$ is an arbitrary quadrangle which satisfies the condition of the exercise, the point $H$ which is obtained as the intersection of the line $EG$ with the side $BC$ must lie between $F$ and $G$. From this and from (1) (analogously, (2)) follows that the point $H$ is unique, i.e., it doesn't depend on the choice of the quadrangle $ABCD$.