# Parallel Lines in Rhombus

### Solution

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: $QT\parallel AP.$

### Acknowledgment

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.

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