, 2006

Point $E$ is the centre of a homothety $h$ which takes circle $\omega_{1}$ to circle $\omega$. The radii $O_{1} D$ and $O B$ of these circles are parallel as both are perpendicular to line $t$. Also, $O_{1} D$ and $O B$ are on the same side of line $E O$, hence $h$ takes $O_{1} D$ to $O B$. Consequently, points $E$, $D$ and $B$ are collinear. Likewise, points $F$, $D$ and $A$ are collinear as well.
Let lines $A E$ and $B F$ intersect at $C$. Since $A F$ and $B E$ are altitudes in triangle $A B C$, their common point $D$ is the orthocentre of this triangle. So $C D$ is perpendicular to $A B$, implying that $C$ lies on line $t$. Note that triangle $A B C$ is acute-angled. We mention the well-known fact that triangles $F E C$ and $A B C$ are similar in ratio $\cos \gamma$, where $\gamma=\angle A C B$. In addition, points $C$, $E$, $D$ and $F$ lie on the circle with diameter $C D$.

Let $P$ be the common point of lines $E F$ and $t$. We are going to prove that $P$ lies on line $A O_{1}$. Denote by $N$ the second common point of circle $\omega_{1}$ and $A C$; this is the point of $\omega_{1}$ diametrically opposite to $D$. By Menelaus' theorem for triangle $D C N$, points $A$, $O_{1}$ and $P$ are collinear if and only if
$$\frac{C A}{A N} \cdot \frac{N O_{1}}{O_{1} D} \cdot \frac{D P}{P C}=1 .$$
Because $N O_{1}=O_{1} D$, this reduces to $C A / A N=C P / P D$. Let line $t$ meet $A B$ at $K$. Then $C A / A N=C K / K D$, so it suffices to show that
$$\begin{equation*} \frac{C P}{P D}=\frac{C K}{K D} . \tag{1} \end{equation*}$$
To verify (1), consider the circumcircle $\Omega$ of triangle $A B C$. Draw its diameter $C U$ through $C$, and let $C U$ meet $A B$ at $V$. Extend $C K$ to meet $\Omega$ at $L$. Since $A B$ is parallel to $U L$, we have $\angle A C U=\angle B C L$. On the other hand $\angle E F C=\angle B A C$, $\angle F E C=\angle A B C$ and $E F / A B=\cos \gamma$, as stated above. So reflection in the bisector of $\angle A C B$ followed by a homothety with centre $C$ and ratio $1 / \cos \gamma$ takes triangle $F E C$ to triangle $A B C$. Consequently, this transformation
takes $C D$ to $C U$, which implies $C P / P D=C V / V U$. Next, we have $K L=K D$, because $D$ is the orthocentre of triangle $A B C$. Hence $C K / K D=C K / K L$. Finally, $C V / V U=C K / K L$ because $A B$ is parallel to $U L$. Relation (1) follows, proving that $P$ lies on line $A O_{1}$. By symmetry, $P$ also lies on line $A O_{2}$ which completes the solution.

We proceed as in the first solution to define a triangle $A B C$ with orthocentre $D$, in which $A F$ and $B E$ are altitudes.
Denote by $M$ the midpoint of $C D$. The quadrilateral $C E D F$ is inscribed in a circle with centre $M$, hence $M C=M E=M D=M F$.

Consider triangles $A B C$ and $O_{1} O_{2} M$. Lines $O_{1} O_{2}$ and $A B$ are parallel, both of them being perpendicular to line $t$. Next, $M O_{1}$ is the line of centres of circles ( $C E F$ ) and $\omega_{1}$ whose common chord is $D E$. Hence $M O_{1}$ bisects $\angle D M E$ which is the external angle at $M$ in the isosceles triangle $C E M$. It follows that $\angle D M O_{1}=\angle D C A$, so that $M O_{1}$ is parallel to $A C$. Likewise, $M O_{2}$ is parallel to $B C$.
Thus the respective sides of triangles $A B C$ and $O_{1} O_{2} M$ are parallel; in addition, these triangles are not congruent. Hence there is a homothety taking $A B C$ to $O_{1} O_{2} M$. The lines $A O_{1}$, $B O_{2}$ and $C M=t$ are concurrent at the centre $Q$ of this homothety.
Finally, apply Pappus' theorem to the triples of collinear points $A, O, B$ and $O_{2}, D, O_{1}$. The theorem implies that the points $A D \cap O O_{2}=F$, $A O_{1} \cap B O_{2}=Q$ and $O O_{1} \cap B D=E$ are collinear. In other words, line $E F$ passes through the common point $Q$ of $A O_{1}$, $B O_{2}$ and $t$.