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Given the parallelogram $ABCD$ in $\mathbb{A}^{2}$ and a point $P$ on the diagonal $AC$. Suppose that $AB \cap DP = \{Q \}$ and $BC \cap DP= \{R \}$. Show that $(D,P,R)=(Q,P,D)$. (For three collinear points $A$, $B$ and $C$, $(A,B,C)$ is defined as $\frac{\vec{AC}}{\vec{AB}}$.)

I already succeeded in providing an analytical proof for this problem(, starting from an affine transformation that maps the parallelogram onto the unit square, which makes it a lot easier to prove). I was wondering however, if it were possible to provide a synthetic proof for this as well? I'm sensing there should be homotheties involved in it; if we take $H$ as the homothety with centre $R$ that maps $B$ onto $C$, then also $H(Q)=D$. I do not know if this is really useful and if it were, I don't know how to go on from here. Any suggestions? Thank you in advance!

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  • $\begingroup$ Are you sure it is $(D,P,R)=(Q,P,D)$? $\endgroup$ – Aqua Jul 29 '18 at 14:41
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Close. Very close! We need a spiral homothety.

Let $\mathcal S$ be a spiral similarity centered at $P$ of $180^\circ$ mapping $A \mapsto C$, $Q \mapsto D$, and $D \mapsto Q$.

$\mathcal S$ preserves $(\cdot, \cdot, \cdot)$.

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