Estonian Math Competitions

Let $K$ and $L$ be the points of intersection of the line $AF$ with lines $BC$ and $DE$, respectively (Fig. 6). Then $$\begin{array}{rl}\angle AGD& =\angle AGF=\angle AGC+\angle CGF=\angle ABC+\angle CAF\\ & =\angle ABK+\angle KAB=\angle CKA=\angle KLD=180^{\circ }-\angle ALD.\end{array}$$ Hence the quadrilateral $AGDL$ is cyclic. Interchanging the roles of points $B$ and $C$, points $D$ and $E$ and also points $G$ and $H$, we can similarly prove that the quadrilateral $AHEL$ is cyclic. Thus the circumcircles of triangles $ADG$ and $AEH$ meet at point $L$, i.e., $P=L$. Point $L$ was chosen on the line $AF$.

Fig. 6

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Alternative solution.

Choose a point $Z$ on the same side of the line $AF$ as point $B$ such that $ZF\parallel BC\parallel DE$ (Fig. 7). Then $\angle ZFB=\angle FBC=\angle FAC=\angle FAB=\angle FCB$. Hence $ZF$ is tangent to the circumcircle of the triangle $ABC$ at $F$, implying that $\angle GFZ=\angle GHF=\angle GHE$. As $\angle GFZ=\angle DFZ=\angle FDE=180^{\circ }-\angle GDE$, we altogether have $\angle GHE=180^{\circ }-\angle GDE$. Consequently, the quadrilateral $DEHG$ is cyclic. Hence $|FD|\cdot |FG|=|FE|\cdot |FH|$, implying that $F$ lies on the radical axis of the circumcircles of triangles $ADG$ and $AEH$. The desired result follows.

Fig. 7