, 2010

Consider three cases $|AB| = |AC|$, $|BC| = |BA|$, and $|CA| = |CB|$.

1. We show that if $|AB| = |AC|$ (Fig. 4), then circle $c$ passes through the circumcenter of $ABC$. Let $O$ be the point at the same side from $AB$ as $C$ that is the intersection of the perpendicular bisector of side $AB$ and circle $c$. Then $\angle OAB = \angle OBA$ and, by inscribed angles theorem,

Fig. 4

$\angle OBA = \angle OAC$. Hence $O$ lies on the bisector of angle $CAB$. Since $|AB| = |AC|$, this angle bisector is also the perpendicular bisector of side $BC$. Consequently, $O$ is the intersection point of the perpendicular bisectors of the sides of triangle $ABC$.

2. For the case $|BC| = |BA|$ (Fig. 5), we show that circle $c$ passes through the orthocenter of triangle $ABC$. Let $E$ be the foot of the altitude of triangle $ABC$ drawn from $B$ and let $H$ be the second intersection point of this altitude with circle $c$ (in the special case with tangency and no intersection, take $H = B$). By the inscribed angles theorem, $\angle EBA = \angle EAH$. Thus $\angle ACB + \angle CAH = \angle CAB + \angle EBA = 90^\circ$ whence $AH \perp BC$. Consequently, $H$ is the orthocenter.

Fig. 5

3. Finally, we show that if $|CA| = |CB|$ (Fig. 6), then circle $c$ passes through the incenter of triangle $ABC$. Let $I$ be the intersection point of the bisector of angle $CAB$ with circle $c$. By the inscribed angles theorem, $\angle CAI = \angle IBA$. Hence $\angle BAI = \angle IBA$ whence $I$ lies on the perpendicular bisector of side $AB$. As $|CA| = |CB|$, this perpendicular bisector is also the bisector of angle $ACB$. Consequently, $I$ is the intersection point of the angle bisectors.

Fig. 6