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\( \triangle\ ABC \ \ \wedge\ \ \left\|\ \begin{array}{ccc}
S\in BC & , & \frac {SB}{SC}=\left(\frac cb\right)^2\\\\\\\\
E\in (AC) & , & SE\ \parallel\ AB\\\\\\\\
F\in (AB) & , & SF\ \parallel\ AC\\\\\\\\
P\in BE & , & CP\ \parallel\ AB\\\\\\\\
R\in CF & , & BR\ \parallel\ AC\end{array}\ \right\|\ \Longrightarrow\ \left\{\ \begin{array}{cc}
\odot & A\in PR\\\\\\\\
\odot & \frac {AP}{AR}=\left(\frac bc\right)^2\\\\\\\\\\\\
\odot & \mathrm {BCPR\ este\ ciclic}\end{array} \) .

Notam T intersectia dintre BR si PC.

Din paralelism se obtine ca \( BR=\frac{c^2}{b} \) si \( CP=\frac{b^2}{c} \).

Notand \( \overline{x} \) si \( \overline{y} \) versorii vectorilor \( \overline{AB} \) respectiv \( \overline{AC} \) atunci

\( \overline{AR}=c\overline{x}-\frac{c^2}{b}\overline{y} \) iar \( \overline{AP}=b\overline{y}-\frac{b^2}{c}\overline{x} \) si de aici A,P si R coliniare.

Deasemenea din asemanare \( \frac{AR}{RP}=\frac{\frac{c^2}{b}}{\frac{c^2}{b}+b}=\frac{c^2}{c^2+b^2} \) iar \( \frac{AP}{PR}=\frac{b^2}{c^2+b^2} \) de unde \( \frac{AP}{AR}=\(\frac{c}{b}\)^2 \)

Din egalitatea \( TB\cdot TR=TC\cdot TP=b^2+c^2 \) se deduce ca BCPR este ciclic.