Consideram un patrulater convex ciclic \( ABCD \) pentru care \( AD>DC \) si \( AB>BC \).
Notam \( \left\{\begin{array}{c}
E\in (AD)\ ,\ DE=DC\\\
F\in (AB)\ ,\ BF=BC\end{array} \) si mijlocul \( G \) al segmentului \( [EF] \). Sa se atate ca \( GB\perp GD \).
Juniori - Iran, 2008
Moderators: Laurian Filip, Filip Chindea, maky, Cosmin Pohoata, Virgil Nicula
Fie \( \omega_1, \omega_2 \) cercurile de centrele \( B,D \) si razele \( BC,CD \) respectiv. Notam \( H \in \omega_1 \cap \omega_2 \), \( M \in GD \cap EH, N \in BG \cap FH \).
Atunci \( \left\| \begin{array}{cc} \angle{EHC} = 180^\circ - \frac {1}{2}\angle{D} \\
\\
\angle{FHC} = 180^\circ - \frac {1}{2}\angle{B} \end{array} \right\| \Rightarrow \angle{EHF} = 360^\circ - \big(360^\circ - \frac {1}{2}(\angle{B} + \angle{D})\big) = 90^\circ \)
Fiindca \( HG \) este mediana avem \( EG = GF = HG \). In patrulaterele \( EDHG \) si \( HGFB \) diagonalele sunt perpendiculare, deci \( \angle{GNH} = \angle{GMH} = 90^\circ \Rightarrow MHNG \) este dreptunghi si \( \angle{DGB} = 90^\circ \).
Atunci \( \left\| \begin{array}{cc} \angle{EHC} = 180^\circ - \frac {1}{2}\angle{D} \\
\\
\angle{FHC} = 180^\circ - \frac {1}{2}\angle{B} \end{array} \right\| \Rightarrow \angle{EHF} = 360^\circ - \big(360^\circ - \frac {1}{2}(\angle{B} + \angle{D})\big) = 90^\circ \)
Fiindca \( HG \) este mediana avem \( EG = GF = HG \). In patrulaterele \( EDHG \) si \( HGFB \) diagonalele sunt perpendiculare, deci \( \angle{GNH} = \angle{GMH} = 90^\circ \Rightarrow MHNG \) este dreptunghi si \( \angle{DGB} = 90^\circ \).
Last edited by Ahiles on Wed Jul 09, 2008 11:10 am, edited 1 time in total.