Estonian Math Competitions

As $\angle ABC=\angle A{P}^{\prime }C$ (see figure), it suffices to show that the quadrilateral $APC{P}^{\prime }$ is a square if and only if $\angle A{P}^{\prime }C=90^{\circ }$.

Since $\angle ABP=\angle CBP$, we have $AP=CP$. Thus $P{P}^{\prime }$ is the perpendicular bisector of the side $AC$. Hence the line $P{P}^{\prime }$ passes through the circumcentre of the triangle $ABC$, i.e., the chord $P{P}^{\prime }$ is a diameter. By Thales' theorem, $\angle PA{P}^{\prime }=\angle PC{P}^{\prime }=90^{\circ }$.

If $\angle A{P}^{\prime }C=90^{\circ }$ then also $\angle APC=180^{\circ }-90^{\circ }=90^{\circ }$. Thus the quadrilateral $APC{P}^{\prime }$ is a rectangle and, by equality of adjacent sides $AP$ and $CP$, a square. On the other hand, if $APC{P}^{\prime }$ is a square then obviously $\angle A{P}^{\prime }C=90^{\circ }$. This completes the proof.



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Alternative solution.

Since $\angle ABC=\angle A{P}^{\prime }C$, it suffices to show that the quadrilateral $APC{P}^{\prime }$ is a square if and only if $\angle A{P}^{\prime }C=90^{\circ }$.

As $\angle ABP=\angle CBP$, we have $AP=CP$. Thus $P{P}^{\prime }$ bisects the line segment $AC$.

If $\angle A{P}^{\prime }C=90^{\circ }$ then, by Thales' theorem, $AC$ is the diameter of the circumcircle of the triangle $ABC$. Thus the chord $P{P}^{\prime }$ bisecting $AC$ passes through the circumcentre of the triangle $ABC$, implying that $AC$ bisects the chord $P{P}^{\prime }$. As $P{P}^{\prime }\perp AC$, the diagonals of the quadrilateral $APC{P}^{\prime }$ are perpendicular and bisect each other. Hence the quadrilateral $APC{P}^{\prime }$ is a rhombus and, because of $\angle A{P}^{\prime }C=90^{\circ }$, a square. On the other hand, if $APC{P}^{\prime }$ is a square then obviously $\angle A{P}^{\prime }C=90^{\circ }$. This completes the proof.