72nd Czech and Slovak Mathematical Olympiad

$$|AB|=|BC|=|CD|=|BR|=|CS|.(1)$$ By construction, the points $R,S$ are obviously different and the midpoint $X$ of the segment $AR$ lies on its perpendicular bisector $BD$ and the midpoint $Y$ of the segment $DS$ lies on its perpendicular bisector $AC$. In the following paragraph we prove that $R$ lies inside the angle $ABC$ and $S$ inside the angle $DCB$, as in our figure. Together, this means that points $A,D,R,S$ lie inside the same half-plane with the boundary line $BC$.



The assumption $|\angle APD|<90^{\circ }$ implies $|\angle APB|>90^{\circ }$, which for the interior point $P$ of the base $AC$ of the isosceles triangle $ABC$ means that $|\angle ABP|<\frac{1}{2}|\angle ABC|$; hence $|\angle ABR|=2\cdot |\angle ABP|<|\angle ABC|$, hence the point $R$ is actually inside the angle $ABC$. Analogously from the inequality $|\angle DPC|>90^{\circ }$ for the interior point $P$ of the base $BD$ of isosceles triangle $DCB$, we conclude that the point $S$ actually lies inside the angle $DCB$.

A further consequence of the inequality $|\angle APD|<90^{\circ }$ is that for the marked interior angles of the right triangles $APX$ and $DPY$, $|\angle XAP|=90^{\circ }-|\angle APD|=|\angle YDP|$, i.e. $|\angle RAC|=|\angle SDB|$.

Let us return to the equalities (1). According to these, the point $B$ is the circumcenter of the triangle $ARC$, which evidently lies in the angle $ABC$. Therefore, according to the inscribed angle theorem $|\angle RBC|=2\cdot |\angle RAC|$. By a similar reasoning about the circumcenter $C$ of the triangle $BSD$ in the angle $BCD$ we obtain $|\angle SCB|=2\cdot |\angle SDB|$. From the last two paragraphs we get the equality $|\angle RBC|=|\angle SCB|$. This, together with (1), leads to the conclusion that (isosceles) triangles $RBC$ and $SCB$ are congruent by the SAS theorem. Hence, their altitudes from the vertices of $R$ and $S$ to the side $BC$ have the same length. This already implies that $BC\parallel RS$.

As in the first solution, we derive (1) and observe that $R$ lies inside the angle $ABC$. From the condition $|\angle APD|<90^{\circ }$ it also follows that $R$ lies in the half plane $ACD$. According to (1), $B$ is the circumcenter of $ARC$, whose central angle $RBA$ with the bisector $BD$ is therefore twice the angle $RCA$. Therefore the three angles $PBA$, $RBP$ and $RCP$ marked in the figure are congruent. Congruence of the last two angles with respect to the previous paragraph already means that the point $R$ does indeed lie on the circumcircle of $BCP$. For the point $S$ the same is true due to the analogous congruence of the angles $PCD$, $SCP$ and $SBP$.



It follows from the proof that the points $B$, $C$, $R$, $S$ lie on one circle, while the points $R$ and $S$ lie in the same half-plane with the boundary line $BC$. Hence the congruence of the angles $BRC$ and $BSC$, which, together with the equality $|BC|=|BR|=|CS|$, means that the isosceles triangles $RBC$ and $SCB$ are congruent. The congruence of their altitudes proves the relation $BC\parallel RS$.

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Alternative solution.

We show that the points $R$ and $S$ lie on the circumcircle of $BCP$. We write the detailed proof only for the point $R$, for the point $S$ the proof is analogous.

As in the first solution, we derive (1) and observe that $R$ lies inside the angle $ABC$. From the condition $|\angle APD|<90^{\circ }$ it also follows that $R$ lies in the half plane $ACD$. According to (1), $B$ is the circumcenter of $ARC$, whose central angle $RBA$ with the bisector $BD$ is therefore twice the angle $RCA$. Therefore the three angles $PBA$, $RBP$ and $RCP$ marked in the figure are congruent. Congruence of the last two angles with respect to the previous paragraph already means that the point $R$ does indeed lie on the circumcircle of $BCP$. For the point $S$ the same is true due to the analogous congruence of the angles $PCD$, $SCP$ and $SBP$.

* Instead of consideration of the congruent triangles $CBR$ and $BCS$, it suffices to state, that the congruent segments $BR$ and $CS$ are symmetrically clustered along the axis of the line segment $BC$.