6-th Czech-Slovak Match

Let us show that $|FA|=|FG|$. We will proceed: from behind. On the extension of $CD$ we take the point $P$ such that $|DP|=|DA|$ and similarly on extension of $DC$ we take the point $Q$ such that $|CQ|=|CA|$. Then, using well known properties of the incircle, we have $$\begin{array}{rl}|PE|& =|PD|+|DE|=|DA|+\frac{|BD|+|CD|-|BC|}{2}=\frac{|BD|+|CD|+|BC|}{2}\\ |QE|& =|QC|+|CE|=|AC|+\frac{|BD|+|CD|-|BC|}{2}=\frac{|BD|+|CD|+|BC|}{2}\end{array}$$ This means that the line $EF$ is the axis of the segment $PQ$. In particular, it means that the circumcenter $O$ of triangle $△APQ$ lies on the line $EF$ as well as on the axes of segments $AP$ and $AQ$. This means that $$\angle DAO=\angle OPD=\angle CQO=\angle OAC$$ So points $O$ and $F$ coincide. Moreover $$\angle AOP=2\angle AQP=\angle AQC+\angle CAQ=180^{\circ }-\angle QCA=\angle ACP$$ Therefore the points $A$, $C$, $F$, $P$ are concyclic. Thus, necessarily $P=G$, and thus $|FP|=|FG|=|FA|$. We have proved what we need. $\square$

We will show that $FA=FG$. Let $H$ be the center of the excircle of triangle $△ACD$ opposite vertex $A$. Then $H$ lies on the angle bisector $AF$. Let $K$ be the point where this excircle touches $CD$. By a standard computation using equal tangents, we see that $CK=(AD+CD-AC)/2$. By a similar computation in triangle $△BCD$, we see that $CE=(BC+CD-BD)/2=CK$. Therefore $E=K$ and $F=H$. Since $F$ is now known to be an excenter, we have that $FC$ is the external angle bisector of $\angle DCA=\angle GCA$. Therefore $$\angle GAF=\angle GCF=90^{\circ }-\frac{1}{2}\angle GCA=90^{\circ }-\frac{1}{2}\angle GFA$$ We conclude that the triangle $△GAF$ is isosceles with $FA=FG$, as desired. $\square$