Czech-Slovak-Polish Match

Let $k$ be the circumcircle of the quadrangle $ABCD$ and $k_1$, $k_2$ the circumcircles of the triangles $PAB$ and $PCD$, respectively. In the interior of the angle $BPC$, consider the half-line $PT$ such that $|\angle BPT|=|\angle BAP|$. Then the hypothesis on $P$ implies that (Fig. 1) $$|\angle TPC|=|\angle BPC|-|\angle BPT|=|\angle BPC|-|\angle BAP|=|\angle PDC|.$$ Thus $PT$ is the common interior tangent of the circles $k_1$ and $k_2$.

Fig. 1

Let us first assume that the sides $AB$ and $CD$ of the given cyclic quadrangle are not parallel. Since the segments $AB$ and $CD$ are common chords of the circles $k_1$, $k$ and $k_2$, $k$, respectively, there exists a unique point $Q$ having the same power with respect to all three circles $k$, $k_1$ and $k_2$. The point $Q$ is the intersection of the three lines $AB$, $DC$ and $PT$. Without loss of generality, we can assume that the point $Q$ is located on the half-line $BA$ beyond the point $A$ (Fig. 1). Then we have $$|\angle QPA|=|\angle PBA|(1)$$ Since $|\angle AEP|=|\angle AFP|=90^{\circ }$, the quadrangle $AEPF$ is cyclic, and $$|\angle FEP|=|\angle FAP|=|\angle DAP|(2)$$ Similarly we see that the quadrangle $DGPF$ is cyclic. It follows that $$|\angle BPC|=|\angle BAP|+|\angle PDC|=|\angle EFP|+|\angle PFG|=|\angle EFG|(3)$$ Since further $|\angle PEQ|=|\angle PGQ|=90^{\circ }$, the quadrangle $QEPG$ is also cyclic and $$|\angle GEP|=|\angle GQP|=|\angle DQP|(4)$$ From the relations (2), (4) and the equality $|\angle DAP|+|\angle QPA|=|\angle QDA|+|\angle DQP|$, we further have $$|\angle FEG|=|\angle FEP|-|\angle GEP|=|\angle DAP|-|\angle DQP|=|\angle QDA|-|\angle QPA|(5)$$ Since the quadrangle $ABCD$ is cyclic, $|\angle QDA|=|\angle QBC|$. From the relations (1) and (5) we thus get $$|\angle FEG|=|\angle QBC|-|\angle PBA|=|\angle PBC|$$ Using finally the relations (3) and (6) we see that the triangles $FEG$ and $PBC$ are similar (as they have two congruent angles).

An analogous argument can be used when the point $Q$ is located on the half-line $AB$ beyond the point $B$. If the lines $AB$ and $CD$ are parallel, then $ABCD$ is an equilateral trapezoid, with bases $AB$ and $CD$. Since the points $E$, $P$, $G$ are collinear and the common interior tangent of the circles $k_1$ and $k_2$ is parallel to both lines $AB$ and $CD$, the triangles $APD$ and $PBC$ are congruent. The similarity of the triangles $EFG$ and $APD$ thus implies also the similarity of the triangles $EFG$ and $PBC$.

This completes the proof.