Balti Tee 2024 · Ülesanne 14

Since $\angle A E H=90^{\circ}=\angle A F H$, we know that $A E H F$ is a cyclic quadrilateral and $A H$ is a diameter of its circumcircle. As $K H \| B C$ and $A H \perp B C$, we have $\angle K H A=90^{\circ}$, so $K H$ is tangent to the circumcircle of $A E H F$.

Denote by $H^{\prime}$ the reflection of $H$ in $K D$ and by $L$ the intersection of lines $K D$ and $H H^{\prime}$ (Fig. 17). Clearly $\angle H L D=90^{\circ}$ and from the equality $\angle K H D=180^{\circ}-\angle H L D$ it follows that $K H$ is also tangent to the circumcircle of $D L H$. From the power of the point $K$ with respect to the circumcircles of $A E H F$ and $D L H$ we obtain $K E \cdot K F=K H^{2}=K L \cdot K D$. Hence points $E, F, D, L$ lie on a common circle.

By a known fact of triangle geometry, reflections of $H$ in the points $D, E$ and $F$ lie on $\omega$. Hence the homothety with center $H$ and ratio 2 maps the circumcircle of triangle $D E F$ to $\omega$. As this homothety maps $L$ to $H^{\prime}$ and $L$ lies on the circumcircle of triangle $D E F$, the point $H^{\prime}$ must lie on $\omega$.

Remark: The problem can be approached using computational methods, namely complex numbers and Cartesian coordinates.

Figure 17