Parallel Lines in Rhombus

Created with GeoGebra

Let $PQ=y+z,\,$ $CQ=x+z,\,$ $CP=x+y.\,$ Then $(O)\,$ is the $C\text{-excircle}\,$ of $\Delta CPQ.\,$ Assume $(O)\,$ touches $CP\,$ at $S.\,$ Then

$\displaystyle \overrightarrow{CS}=\frac{(x+y+z)\overrightarrow{CP}}{x+y}\,$ and $\displaystyle \overrightarrow{CO}=\frac{(x+z)\overrightarrow{CP}+(x+y)\overrightarrow{CQ}}{2x}$

from which $\displaystyle \overrightarrow{CA}=\frac{(x+z)\overrightarrow{CP}+(x+y)\overrightarrow{CQ}}{x}.$

But $\overrightarrow{AT}=-\overrightarrow{CS},\,$ so that $\overrightarrow{AC}+\overrightarrow{CT}=-\overrightarrow{CS},\,$ $\overrightarrow{CT}=\overrightarrow{CA}-\overrightarrow{CS},\,$ $\overrightarrow{CQ}+\overrightarrow{QT}=\overrightarrow{CA}-\overrightarrow{CS},$

$\displaystyle \begin{align} \overrightarrow{QT}&=\overrightarrow{CA}-\overrightarrow{CS}-\overrightarrow{CQ}\\ &=\frac{(x+z)\overrightarrow{CP}+(x+y)\overrightarrow{CQ}}{x}-\frac{(x+y+z)\overrightarrow{CP}}{x+y}-\overrightarrow{CQ}, \end{align}$

which simplifies to $\displaystyle \overrightarrow{QT}=\frac{y}{x(x+y)}\cdot [z\overrightarrow{CP}+(x+y)\overrightarrow{CQ}].$

On the other hand,

$\displaystyle\begin{align} \overrightarrow{PA}&=\overrightarrow{CA}-\overrightarrow{CP}\\ &=\frac{(x+z)\overrightarrow{CP}+(x+y)\overrightarrow{CQ}}{x}-\overrightarrow{CP}\\ &=\frac{1}{x}\cdot[z\overrightarrow{CP}+(x+y)\overrightarrow{CQ}] \end{align}$

which, in particular, proves the claim: $QT\parallel AP.$

This problem has been kindly communicated to me by Leo Giugiuc, along with his solution. The problem by Rubén Auqui (Dario) was originally posted at the Peru Geometrico facebook group.