Here's my solution: Lemma. Let $\omega$ be a circle with diameter $AB$ and center $O$, let $C$ and $D$ two points on $\omega$, and let $K:= AD\cap CB$. Then $K$ lies on the circle that is tangent to $OC$ and $OD$ through $C$ and $D$ (which we call $\Omega$), respectively. Moreover, $K$ is the point on $\Omega$ resulting from rotating the line parallel to $AB$ and passing through the center of $\Omega$ by $90^\circ$ towards $D$. Proof. (We use directed angles.) Define $J:= AC\cap BD$. We note that $\angle JDK = \angle KCJ = 90^\circ$, so $JK$ is a diameter of $\Omega$. From angle chasing, we have $\angle DCJ = \angle DCA = \angle DBA = \angle DBO = \angle ODB = \angle ODJ$. This shows that $DO$ is tangent to the circumcircle of $DCJ$. By symmetry, we conclude that $CO$ is also tangent to this circumcircle. But then this implies that this circumcircle is $\Omega$. As for the second part, we note that $\angle DBA = \angle DCJ = \angle DKJ = 90^\circ - \angle KJD$, so that the angle between $JK$ and $AB$ is $90^\circ$, and we are done. Now we prove the main theorem: Let $\omega_A, \omega_B, \omega_C$ be the circles with diameter $A_1 A_2, B_1 B_2, C_1 C_2$, and let $O_A, O_B, O_C$ be their diameters, respectively. Let $D, E, F$ be the tangency points of $\omega_B$ and $\omega_C$, $\omega_C$ and $\omega_A$, $\omega_A$ and $\omega_B$, respectively; note that $D, E, F$ exist by assumption. Let $\omega_{DEF}$ be the circumcircle of $DEF$, and let $P$ be the point obtained by rotating the line passing through the center of $\omega_{DEF}$ and parallel to any of $A_1 A_2, B_1 B_2, C_1 C_2$ by $90^\circ$. Then $A_1 B_2\cap C_1 A_2 = A_1 E\cap A_2 F$. By the Lemma proved earlier, this intersection is equal to the rotation of $A_1 A_2$ by $90^\circ$ in the circle tangent to $O_A E, O_A F$ through $E, F$, respectively. But this circle is equal to $\omega_{DEF}$, so this intersection is equal to $P$. Applying the previous proof cyclically, we show that $A_1 B_2, B_1 C_2, C_1 A_2$ all pass through $P$, and we are done.