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O frumoasa proprietate a unui patrulater inscriptibil.
Posted: Fri Apr 24, 2009 3:37 pm
by Virgil Nicula
Fie un patrulater convex inscriptiibil \( ABCD \) pentru care notam : \( E\in AB\cap CD \) si \( F\in BC\cap AD \) ; mijloacele \( X \) , \( Y \) pentru \( [AC] \) , \( [BD] \) respectiv; intersectia \( P \) a bisectoarelor pentru \( \angle AED \) , \( \angle CFD \) . Sa se arate ca \( P\in XY \) si \( \frac {PX}{PY}=\frac {AC}{BD} \) .
Posted: Sat Apr 25, 2009 1:13 pm
by Mateescu Constantin
Notam \( \left\|\begin{array}{cccc}
& T & \in & FP & \cap & CD \\\\
& R & \in & FP & \cap & AB \\\\
& U & \in & EP & \cap & AD \\\\
& S & \in & EP & \cap & BC \end{array}\right\| \) . Se demonstreaza destul de usor ca patrulaterul \( TURS \) este romb \( (*) \) .
Avem: \( \left\|\ \begin{array}{ccc}
\triangle FCA \sim \triangle FDB\ \mbox{(U.U.)}\ \ \Longrightarrow\ \frac{FA}{FB}=\frac{FC}{FD}=\frac{AC}{BD} \\\\\\\\
\mbox{T.Bis.}\ \ \begin{array}
& \normal{\nearrow} & \triangle FAB & \Longrightarrow & \frac{FA}{FB}=\frac{AR}{RB}\\\\\\\\
& \normal{\searrow} & \triangle FCD & \Longrightarrow & \frac{FC}{FD}=\frac{CT}{TD}
\end{array}\end{array}\ \right\|\ \Longrightarrow\ \frac{AR}{RB}=\frac{CT}{TD}=\frac{AC}{BD}=k\ (**) \)
Din relatiile vectoriale avem: \( \left\|\ \begin{array}{ccc}
\vec{XP} & = & \frac{\vec{XT}+\vec{XR}}{2}\ (*)\\\\\\\\
\vec{XT} & = & \frac{\vec{AT}+\vec{CT}}{2}\\\\\\\\
\vec{XR} & = & \frac{\vec{AR}+\vec{CR}}{2}
\end{array}\ \right\|\ (1) \) . Din \( (**)\ \Longrightarrow\ \left\|\ \begin{array}
\vec{AT} & = & \frac{\vec{AC}+k\vec{AD}}{k+1}\\\\\
\vec{CT} & = & \frac{k}{k+1}\vec{CD} \\\\\\
\vec{AR} & = & \frac{k}{k+1}\vec{AB} \\\\\\\
\vec{CR} & = & \frac{\vec{CA}+k\vec{CD}}{k+1}\end{array}\ \right\|\ (2) \)
\( \Longrightarrow^{(1)}\ \vec{XP}=\frac{\vec{AT}+\vec{CT}+\vec{AR}+\vec{CR}}{4}\ =^{(2)}\ \frac{\frac{\vec{AC}}{k+1}+\frac{k\vec{AD}}{k+1}+\frac{k\vec{CD}}{k+1}+\frac{k\vec{AB}}{k+1}+\frac{\vec{CA}}{k+1}+\frac{k\vec{CB}}{k+1}}{4} \)
\( \Longleftrightarrow\ \overline{\underline{\left\|\ \vec{XP}=\frac{k(\vec{AD}+\vec{CD}+\vec{AB}+\vec{CB})}{4(k+1)}\ \right\| \)
In mod analog exprimam vectorul \( \vec{PY} \) si vom obtine: \( \underline{\overline{\left\|\ \vec{PY}=\frac{\vec{AD}+\vec{CD}+\vec{AB}+\vec{CB}}{4(k+1)}\ \right\| \) .
Din ultimele doua relatii deducem ca vectorii \( \vec{XP} \) si \( \vec{PY} \) sunt coliniari deci punctele \( X \), \( P \) si \( Y \) sunt coliniare.
Tot de aici rezulta ca \( \frac{PX}{PY}=k\ \Longleftrightarrow^{(**)}\ \frac{PX}{PY}=\frac{AC}{BD} \) .