IRL ABooklet

Let $\beta = \angle ABC$, $\gamma = \angle BCA$ and let $\Omega_B$ denote the circle through $A$ and $B$ and $\Omega_C$ the circle through $A$ and $C$. The assumptions about the two circles mean that for any point $D$ on $\Omega_B$ which is on the arc $AB$ that is outside $\triangle ABC$ we have $\angle BDA = 90^\circ - \gamma$. Similarly, for $E$ on $\Omega_C$ on the correct arc $AC$ we have $\angle AEC = 90^\circ - \beta$.

Let $T$ be a point inside triangle $ABC$ on the tangent to $\Omega_B$ at $A$. By the Alternate Segment Theorem, we have $\angle BAT = \angle BDA = 90^\circ - \gamma$. Hence,
$$\angle TAC = \angle BAC - \angle BAT = (180^\circ - \beta - \gamma) - (90^\circ - \gamma) = 90^\circ - \beta = \angle AEC.$$
The converse of the Alternate Segment Theorem implies now that $TA$ is tangent to $\Omega_C$ at $A$, i.e. $\Omega_B$ and $\Omega_C$ touch each other at $A$. When $\angle BAC = 90^\circ$, we have $\beta + \gamma = 90^\circ$ and so $\beta = 90^\circ - \gamma = \angle BDA$ as well as $\gamma = 90^\circ - \beta = \angle AEC$. The converse of the Alternate Segment Theorem implies then that $BC$ is tangent to both circles.

We use notation from Solution 1 and let $O_B$ and $O_C$ be the centres of $\Omega_B$ and $\Omega_C$, respectively. The assumptions about the angle in the segment external to the triangle imply that the central angle $\angle AO_B B$ equals $180^\circ - 2\gamma$ and that $\angle AO_C C = 180^\circ - 2\beta$. These equations imply $\angle O_B AB = \gamma$ and $\angle O_C AC = \beta$. Because $\angle BAC = 180^\circ - \beta - \gamma$ we see now that $O_A, O_B$ and $A$ are collinear, hence the two circles touch each other at $A$.

For the right angled case, see the end of Solution 1.