IRL ABooklet

Let $M$ be the centre of $\Omega_3$. Then, $A$, $Q$, $M$ are collinear as are $C$, $R$, $M$. Let $H$ be the intersection point of the lines $QR$ and $AD$.

Because $|BR| = |CR|$ and $|DM| = |CM|$, we have $\angle MDC = \angle MCD = \angle RCB = \angle RBC$, hence $DM$ is parallel to $BR$. Using that $R$, $B$, $Q$, $H$ are collinear and $|DM| = |AM|$ we now get $\angle AHQ = \angle ADM = \angle MAD = \angle QAH$ hence triangle $AHQ$ is isosceles with $|QH| = |QA|$ and so $H$ is on $\Omega_1$.

Alternatively, we may consider $\triangle HBA$, which has a right angle at $A$. Because the centre, $Q$, of $\Omega_1$ is on $HB$, this implies that $H$ must be on $\Omega_1$.

Draw tangents at $A$ and $C$ to $\Omega_3$ and let them intersect at $T$. The radical axis of two circles that are tangent to each other is their common tangent. Therefore, $T$ is the radical centre of $\Omega_1$, $\Omega_2$ and $\Omega_3$, which implies that $BT$ is the common tangent of $\Omega_1$ and $\Omega_2$. Hence, $TB$ is perpendicular to $QR$. Let $E$ be the second intersection point of the line $AB$ with $\Omega_3$, $H$ the intersection point of the lines $QR$ and $AD$, and $F$ the intersection point of $TB$ and $DE$. From $|TA| = |TB| = |TC|$ we obtain
$$\begin{align*} \alpha &:= \angle TAC = \angle TCA \\ \beta &:= \angle TAB = \angle TBA \\ \gamma &:= \angle TCB = \angle TBC = \angle FBD. \end{align*}$$
Considering the angle sum in $\triangle ABC$ we see that $180^\circ = \angle BAC + \angle ABC + \angle BCA = (\beta - \alpha) + (\beta + \gamma) + (\gamma - \alpha) = 2(\gamma + \beta - \alpha)$, hence $\gamma + \beta - \alpha = 90^\circ$.
Considering the external angle at $B$ in $\triangle BCE$, we obtain $\angle BCE + \angle AEC = \angle ABC$, i.e. $\angle BCE = \angle ABC - \angle AEC = \gamma + \beta - \alpha = 90^\circ$.
The Alternate Segment Theorem applied to circle $\Omega_3$ gives us
$$\begin{align*} \angle AEC &= \angle TAC = \alpha \\ \angle ADE &= \angle TAB = \beta \\ \angle DEC &= \angle TCB = \gamma. \end{align*}$$
Similarly, when we consider the external angle at $B$ in $\triangle BAD$ we can use $\angle ADC = \angle AEC = \alpha$ to obtain $\angle BAD = 90^\circ$.
We can now conclude in two different ways. One way is to consider the triangles $\triangle DFB$ and $\triangle DCE$ which have an angle in common at $D$. Because $\angle FBD = \angle DEC = \gamma$, we see that $\angle DFB = \angle DCE = 90^\circ$. Hence $TB$ is perpendicular to $DE$ and so $DE \parallel QR$. This implies that
$$\angle AHR = \angle ADE = \beta = \angle TAB,$$
hence $H$ is on $\Omega_1$ by the converse of the Alternate Segment Theorem.

Alternatively, we may consider $\triangle HBA$, which has a right angle at $A$. Because the centre, $Q$, of $\Omega_1$ is on $HB$, this implies that $H$ must be on $\Omega_1$.

We use the diagram and notations from Solution 2 and let $M$ be the centre of $\Omega_3$. Then the homothety of centre $A$ sending $\Omega_1$ to $\Omega_3$ sends $QB$ to $ME$ and the homothety of centre $C$ sending $\Omega_2$ to $\Omega_3$ sends $RB$ to $MD$. Since $Q$, $B$, $R$ are collinear, it follows that $D$, $M$, $E$ are collinear hence $\angle BAD = 90^\circ$. We conclude as in Solution 2.

Extend $AQ$ to the diameter $AL$ of circle $\Omega_1$ and let $K$ be the second intersection point of $AC$ with $\Omega_1$. Let $M$ be the centre of $\Omega_3$.

Consider the angle between $AC$ and the common tangent to $\Omega_1$ and $\Omega_3$ at $A$. The Alternate Segment Theorem for these circles then yields
$$\angle ADC = \angle ALK. \qquad (2)$$
Note that $\angle QBD = \angle RBC = \angle BCR$ as well as $\angle QAB = \angle QBA$. Because $\angle MAC = \angle ACM$, we get
$$\angle BCR + \angle BCA = \angle QBA + \angle BAC. \qquad (3)$$
Considering the external angle of triangle $ABC$ at $B$, we get
$$\angle BAC + \angle BCA = \angle QBA + \angle QBD. \qquad (4)$$
Subtracting (3) from (4), we obtain now $\angle BAC = \angle QBD$. Together with $\angle QAB = \angle QBA$ we now see that
$$\angle LAK = \angle QAB + \angle BAC = \angle QBA + \angle QBD = \angle ABD.$$
If we combine this with (2), we get
$$\angle DAB = 180^\circ - \angle ADC - \angle ABD = 180^\circ - \angle ALK - \angle LAK = \angle AKL = 90^\circ.$$
If $H$ denotes the second intersection point of $AD$ and $\Omega_1$, we have $\angle HAB = \angle DAB = 90^\circ$, i.e. $HB$ is a diameter of $\Omega_1$ and so $H$ is the intersection point of $QR$ and $AD$.

Let $H$ be the intersection point of the lines $QR$ and $AD$. Lines $AQ$ and $CR$ meet at the centre $M$ of circle $\Omega_3$.

From $\triangle BRC$ we see that $\angle QRM = \angle BCR + \angle RBC = 2\angle RBC = 2\angle HBD$. The central angle $\angle AMC$ stands on the same arc of $\Omega_3$ as $\angle ADC$, hence $\angle QMR = 2\angle HDB$. Considering external angles of triangles $QMR$ and $HDB$ we obtain
$$\angle AQR = \angle QMR + \angle QRM = 2\angle HDB + 2\angle HBD = 2\angle AHB$$
and this implies that $H$ is on $\Omega_1$.