Solution:

Let the internal and external bisectors of angle $BAC$ intersect the line $BC$ respectively at points $D$ and (possibly at infinity) $D_{1}$. The common tangents intersect at the center $T$ of the positive homothety $\mathscr{H}$ which maps the inscribed circle $\omega_{a}$ to the circumscribed circle $\Omega$. If $T$ is a point at infinity, $\mathscr{H}$ is a translation, and the rest of the proof is the same.

Lemma. Let an arbitrary line $p$ through $D_{1}$ intersect the circle $\Omega$ at points $L$ and $K$. The tangents at $L$ and $K$ to $\Omega$ intersect the line $BC$ respectively at points $P$ and $Q$. Then $\text{PAB= CAQ}$.

Proof. Denote $\text{BAC= , CBA= , ACB= , PAB=x}$ and $\text{CAQ=y}$.

If $D_{1}$ is a point at infinity, the claim is trivial by symmetry. If not, from $\triangle PBL \sim \triangle PLC$ it follows that $\frac{PB}{PL}=\frac{PL}{PC}=\frac{LB}{LC}$ and from this $\frac{PB}{PC}=\left(\frac{LB}{LC}\right)^{2}$. Similarly $\frac{QB}{QC}=\left(\frac{KB}{KC}\right)^{2}$. Since $\frac{LB}{LC} \cdot \frac{KB}{KC}=\frac{|KLB|}{|KLC|}=\frac{D_{1}B}{D_{1}C}=\frac{AB}{AC}$, we obtain $\frac{PB}{PC} \cdot \frac{QB}{QC}=\left(\frac{AB}{AC}\right)^{2}$. Since $\frac{PB}{PC}=\frac{PB}{PA} \cdot \frac{PA}{PC}=\frac{\sin x}{\sin \beta} \cdot \frac{\sin \gamma}{\sin (\alpha+x)}$ and $\frac{QB}{QC}=\frac{QB}{QA} \cdot \frac{QA}{QC}=\frac{\sin (\alpha+y)}{\sin \beta} \cdot \frac{\sin \gamma}{\sin y}$, multiplying gives $\left(\frac{\sin \gamma}{\sin \beta}\right)^{2} \cdot \frac{\sin (\alpha+y) / \sin y}{\sin (\alpha+x) / \sin x}=\left(\frac{AC}{AB}\right)^{2}=\left(\frac{\sin \gamma}{\sin \beta}\right)^{2}$, whence $\sin \alpha \operatorname{ctg} y+\cos \alpha=\frac{\sin (\alpha+y)}{\sin y}=\frac{\sin (\alpha+x)}{\sin x}=\sin \alpha \operatorname{ctg} x+\cos \alpha$, i.e. $x=y$.

If $K$ and $L$ are the points of tangency of the common tangents with $\Omega$, it remains to show that point $D_{1}$ lies on the line $KL$, i.e. on the polar of point $T$ with respect to $\Omega$. By the theorem on pole and polar, it suffices to prove that $T$ lies on the polar $d$ of point $D_{1}$ with respect to $\Omega$.

Denote by $N$ the midpoint of the arc $BAC$ of the circle $\Omega$. The image of point $D$ under the homothety $\mathscr{H}$ is the intersection $S$ of the tangents to $\Omega$ at points $A$ and $N$, so point $T$ lies on the line $DS$. On the other hand, point $D$ is on the polar $d$ because the quadruple $\left(B, C ; D_{1}, D\right)$ is harmonic, and point $S$ is

also on $d$ because the polar of point $S$ with respect to $\Omega$, which is the line $AN$, contains point $D_{1}$. Therefore, the lines $DS$ and $d$ coincide, which completes the proof.

Solution:

Second solution. Let the common tangents touch the circle $\Omega$ at points $K$ and $L$, where $LP$ is the tangent closer to vertex $B$. Denote by $M$ the midpoint of the arc $BC$ not containing point $A$, and by $O$ and $I_{a}$ respectively the centers of the circumscribed circle and the circle inscribed opposite $A$.

Since $\text{LPI a =90 + 1 2 LPC}$ and $\text{LAI a = LAM= 1 2 LOM= 1 2 LPD 1 =90 - 1 2 LPC}$, it follows that $\text{LPI a + LAI a =180}$, so the quadrilateral $ALPI_{a}$ is cyclic. Similarly, the quadrilateral $AKQI_{a}$ is also cyclic. Now we have $\text{PAI a = PLI a = QKI a = QAI a}$, since the angles $PLI_{a}$ and $QKI_{a}$ are symmetric with respect to the line $OI_{a}$, and from this it follows that $\text{PAB= QAC}$.