Czech and Slovak Mathematical Olympiad

Since the line segments $AB'$, $AC'$ have the same length $|AB'| + |AC'|$, $AB'C'$ is an isosceles triangle with base $B'C'$. It means that the perpendicular bisector of $B'C'$ coincides with the bisector of the angle $BAC$. Let $S \neq A$ be the intersection of this bisector with the circumcircle of $ABC$. If we prove that $S$ is a circumcenter of $AB'C'$ we will be finished. Since $S$ lies on the perpendicular bisector of $B'C'$, we have $|SB'| = |SC'|$. So, it remains to prove that $|SA| = |SC'|$.

From congruence of inscribed angles $SAB$ and $SAC$ it follows that $S$ is the midpoint of the arc $BC$, and therefore $|BS| = |CS|$. From the cyclic quadrilateral $ABSC$ we have $|\angle ACS| = 180^\circ - |\angle SBA| = |\angle C'BS|$. Together with equality $|CA| = |BC'|$ we get that triangles $SAC$ and $SC'B$ are congruent by condition $SAS$ and therefore $|SA| = |SC'|$.

Figure 1

Let us define the point $S$ as in the first solution. This time we verify the desired equality $|SA| = |SC'|$ by showing that $S$ lies on the perpendicular bisector of $AC'$.

In the special case where $|AB| = |AC|$ holds, the midpoint of $AC'$ is $B$ (according to the construction of $C'$); therefore it suffices to verify that the angle $ABS$ is right. However, this follows from the fact that the cyclic quadrilateral $ABSC$ is then composed of two identical triangles $ABS$ and $ACS$, so the angles at their opposite vertices $B$ and $C$ are identical and therefore right.

When $|AB| \neq |AC|$, we can without loss of generality assume that $|AB| > |AC|$ as in figure 2. Here $P$ and $Q$ denote the perpendicular projections of $S$ onto lines $AB$ and $AC$, respectively. Thanks to our assumption $|AB| > |AC|$ the point $P$ lies inside the segment $AB$, while the point $Q$ lies on the opposite ray to the ray $CA$. We prove that $P$ is the midpoint of $AC'$.

BPS and CQS, also the angles PSB and QSC have the same size. In addition, we have $|PS| = |QS|$, since $S$ lies on the angle bisector of $C'AB'$. We thus obtain that triangles $PBS$ and $QCS$ are congruent according to condition ASA. Hence, equality $|BP| = |CQ|$ follows. Moreover, from the identical rectangular triangles $ASP$ and $ASQ$ we also have $|AP| = |AQ|$, so together it yields
$$|AP| = |AQ| = |AC| + |CQ| = |C'B| + |BP| = |C'P|.$$
Thus, $P$ is indeed the midpoint of $AC'$, and the proof is complete.

Figure 2

This time we denote by $S$ the circumcenter of $AB'C'$ and we prove that points $A$, $B$, $S$, and $C$ lie on one circle. According to the introduction of the first solution, we know that $AB'C'$ is an isosceles triangle with base $B'C'$, hence its circumcenter $S$ lies on the angle bisector of $C'AB'$, that is $BAC$. The points $B$ and $C$ therefore lie in the opposite half-planes with the boundary line $AS$, therefore it is sufficient to verify $|\angle ABS| = 180^\circ - |\angle ACS|$.

By equalities $|AC'| = |AB'|$ and $|AS| = |C'S| = |B'S|$, the triangles $C'AS$ and $AB'S$ are isosceles and congruent. Thus, if we rotate triangle $C'AS$ around $S$ about the oriented angle $C'SA$ we obtain triangle $AB'S$. Since $B$ lies on $C'A$, $C$ lies on $AB'$ and at the same time $|C'B| = |AC|$, this rotation maps $B$ onto $C$ and thus the angle $C'BS$ to the angle $ACS$. Therefore, $|\angle C'BS| = |\angle ACS|$, which already follows $|\angle ABS| = 180^\circ - |\angle C'BS| = 180^\circ - |\angle ACS|$ as we promised to show.

Figure 3