Balti Tee 1997 · Ülesanne 13

Solution:

Let $O, H', G'$ be the circumcentres of the triangles $B D F$, $B C F$ and $C D F$, respectively (see Figure 6). Then $O, G$ and $G'$ lie on the perpendicular bisector of the segment $D F$, while $O, H$ and $H'$ lie on the perpendicular bisector of the segment $B F$. Moreover, $G$ and $H'$ lie on the perpendicular bisector of $B C$, $O$ lies on the perpendicular bisector of $B D$, $H$ and $G'$ lie on the perpendicular bisector of $C D$ and $C$ is the midpoint of $B D$. Hence $H'$ and $G'$ are symmetric to $H$ and $G$, respectively, relative to point $O$. Hence triangles $O G H'$ and $O G' H$ are congruent, and $G H G' H'$ is a parallelogram.

Since $C F$ is the common side of triangles $B C F$ and $C D F$, the line $G' H'$ connecting their circumcentres is perpendicular to $C F$. Therefore $G H$ is also perpendicular to $C F$.

Figure 6

Alternative solution.

Note that the diagonals of a quadrangle $X Y Z W$ are perpendicular to each other if and only if $|X Y|^{2}-|Z Y|^{2}=|X W|^{2}-|Z W|^{2}$. Applying this to the quadrangle $G F H C$ it is sufficient to prove that $|G F|^{2}-|H F|^{2}=|G C|^{2}-|H C|^{2}$. Denote $|A B|=|B C|=|C D|=|D E|=a$, $\angle G A C=\alpha$ and $\angle H E C=\beta$, and let $R_{1}, R_{2}$ be the circumradii of triangles $A D F$ and $B E F$, respectively (see Figure 7). Applying the cosine law to triangles $C G A$ and $C H E$, we have

$$|G C|^{2}=R_{1}^{2}+4 a^{2}-4 a R_{1} \cos \alpha$$

and

$$|H C|^{2}=R_{2}^{2}+4 a^{2}-4 a R_{2} \cos \beta.$$

Together with $\cos \alpha=\frac{3 a}{2 R_{1}}$ and $\cos \beta=\frac{3 a}{2 R_{2}}$ this yields $|G C|^{2}-|H C|^{2}=R_{1}^{2}-R_{2}^{2}$. Since $|G F|=R_{1}$ and $|H F|=R_{2}$, we also have $|G F|^{2}-|H F|^{2}=R_{1}^{2}-R_{2}^{2}$.

Figure 7

Another solution.

We shall use the following fact that can easily be derived from the properties of the power of a point: Let a line $s$ intersect two circles at points $K, L$ and $M, N$, respectively, and let these circles intersect each other at $P$ and $Q$. A point $X$ on the line $s$ lies also on the line $P Q$ (i.e. is the intersection point of the lines $s$ and $P Q$) if and only if $|K X| \cdot|L X|=|M X| \cdot|N X|$.

The line $A E$ intersects the circumcircles of triangles $A D F$ and $B E F$ at $A, D$ and $B, E$, respectively. Since point $C$ lies on line $A E$ and $|A C| \cdot|D C|=|B C| \cdot|E C|$, then line $C F$ passes through the second intersection point of these circles (see Figure 8) and hence is perpendicular to the segment $G H$ connecting the centres of these circles.

Figure 8