Czech and Slovak Mathematical Olympiad

Let us use the standard notation of the inner angles of the triangle $ABC$, further let $I$ be the incenter and $I_a$ be the excenter (of the excircle opposite $A$), and let $D$ and $E$ be in order the touching points of the thought circles. Since the bisectors $BI$ and $BI_a$ of the supplementary angles are perpendicular to each other (as well as $CI$ and $CI_a$), the points $B$, $C$, $I$, and $I_a$ lie on the circle with the diameter $II_a$.

Thus $D$ and $E$, the orthogonal projections of $I$ and $I_a$ onto the secant $BC$, are point reflections of each other with respect to the center of $BC$.

The right triangles $BID$ and $I_aBE$ are obviously similar and
$$|BD| : |ID| = |I_aE| : |BE| \quad \text{or} \quad |BD| \cdot |BE| = |ID| \cdot |I_aE|$$
considering the mentioned point reflection also
$$|BD| + |BE| = |BD| + |CD| = |BC| = r + r_a = |ID| + |I_aE|.$$

Fig. 1

The two equations imply that the pairs ($|ID|$, $|EI_a|$) and ($|BD|$, $|BE|$) are roots of the same quadratic equation, that is $|ID| = |BD|$ or $|ID| = |BE|$.

$|ID| = |BD|$ means the right-angled triangle $BID$ is isosceles, which is $\beta = 90^\circ$.

Similarly, if $|ID| = |BE|$ that is $|ID| = |CD|$ ($D$ and $E$ are point reflections in the mentioned reflection) means the right triangle $CID$ is isosceles, that is $\gamma = 90^\circ$.

In both cases the triangle $ABC$ is right-angled.