The Problems of Ukrainian Authors

Let $P_1$, $P_2$ be respectively the feet of perpendiculars from $P$ to lines $BC$ and $AD$, $M_1$, $M_2$ respectively are midpoints of sides $BC$ and $AD$ respectively, the excircle of the triangle $EBC$ touches the side $BC$ at the point $T^{\prime }$. $T^{\prime }C=B{T}_1\Rightarrow \frac{B{T}_1}{T_{1}C}=\frac{C{T}^{\prime }}{T^{\prime }B}$. $ABCD$ is a cyclic quadrilateral then $△EBC\sim △EAD\Rightarrow \frac{B{T}_1}{T_{1}C}=\frac{C{T}^{\prime }}{T^{\prime }B}=\frac{A{T}_2}{T_{2}D}$ (Fig.22).

Moreover $△PBC\sim △PAD\Rightarrow \frac{BP}{P_{1}C}=\frac{A{P}_2}{P_{2}D}$. Since $M_1$, $M_2$ are the midpoints of segments $BC$ and $AD$ respectively then $\frac{T_1{M}_1}{M_1{P}_1}=\frac{T_2{M}_2}{M_2{P}_2}$.

Let the point $O_1$ be located on the segment $PQ$ such that $\frac{Q{O}_1}{O_{1}P}=\frac{T_1{M}_1}{M_1{P}_1}$. Then the projections of this point on the segments $BC$ and $AD$ respectively coincide with points $M_1$ and $M_2\Rightarrow O$ coincides with $O_1$. Then points $O$, $P$ and $Q$ are collinear.

Fig.22