Estonian Mathematical Olympiad

Since triangle $BAC$ is isosceles (Fig. 25), its base angles are equal; denote $\angle ABC=\angle BCA=\beta$. Since the sum of the interior angles of triangle $BAC$ is $180^{\circ }$, we have $\angle BAC=180^{\circ }-2\beta$. Then $\angle DAF=\frac{180^{\circ }-2\beta }{2}=90^{\circ }-\beta$ because $\angle DAF=\frac{1}{2}\angle BAC$. Since $FA$ is perpendicular to line $BA$, $$\angle BAD=\angle BAF-\angle DAF=90^{\circ }-(90^{\circ }-\beta )=\beta .$$ Thus triangle $ADB$ is isosceles, because angles $DBA$ and $BAD$ are equal. Consequently $AD=BD=7$ cm. Fig. 25 As points $D$ and $E$ are symmetric w.r.t. the perpendicular bisector of $BC$ and so are points $B$ and $C$, we also have $AE=CE=7$ cm. Hence the triangle $AEC$ is isosceles, too. From triangle $BAF$ we now get $$\angle AFB=\angle AFD=180^{\circ }-90^{\circ }-\beta =90^{\circ }-\beta .$$ Thus triangle $DAF$ is isosceles, because angles $DAF$ and $DFA$ are equal. Hence $DF=7$ cm. Therefore $DF=EC$. As the altitude driven from the apex of an isosceles triangle bisects the base, $G$ must be the midpoint of $AC$. Since $H$ is a midpoint of $EF$ and $DF=EC$ (Figures 26 and 27 depict two possible cases), $H$ is also a midpoint of $DC$. In conclusion $HG$ is a midsegment of triangle $ACD$ parallel to side $AD$. So $HG=\frac{1}{2}AD=3.5$ cm.