Thứ Bảy, 13 tháng 9, 2014

42-Tangential quadrilateral in quadrilateral

Let ABCD be a any quadrilateral. E,F,G,H are intersection of angle bisector of four vertex \angle A, \angle B, \angle C, \angle D (Show in the figure attachment. M,N,P,Q are intersection of four line through E,F,G,H and pependicular with AB, AD, DC,CB (show in the figure).

http://www.geogebratube.org/student/m79112 

1-Prove that MNPQ are tangential quadrilateral.

2-Prove that if ABCD is cyclic, then MNPQ is a point T, and center of circle (ABCD),T and intersection of AC,BD are collinear


Solution by: Luis González :P :roll: 



1) Label \widehat{A}, \widehat{B},\widehat{C},\widehat{D} the angles of ABCD. Then \widehat{FAD}=90^{\circ}-\tfrac{1}{2}\widehat{A} and \widehat{FDA}=90^{\circ}-\tfrac{1}{2}\widehat{D} \Longrightarrow \widehat{EFG}=\tfrac{1}{2}(\widehat{A}+\widehat{D}). Similarly we have \widehat{EHG}=\tfrac{1}{2}(\widehat{C}+\widehat{B}) \Longrightarrow \widehat{EFG}+\widehat{EHG}=\tfrac{1}{2}(\widehat{A}+\widehat{B}+\widehat{C}+\widehat{D})=180^{\circ} \Longrightarrow EFGH is cyclic and let T be its circumcenter.

Since PF \perp AD, PE \perp AB, we get \widehat{PFA}=\widehat{PEA}=\tfrac{\widehat{A}}{2} \Longrightarrow \triangle PEF is P-isosceles \Longrightarrow PT is perpendicular bisector of \overline{EF} \Longrightarrow PT bisects \widehat{NPQ}.Similarly, QT, MT, NT bisect \widehat{PQM}, \widehat{QMN}, \widehat{MNP} \Longrightarrow MNPQ is tangential with incenter T. 

2) Let X \equiv BC \cap AD and WLOG assume that E is incenter of \triangle XAB and Gis X-excenter of \triangle XDC. Then \widehat{DGE} \equiv \widehat{DGX}=\tfrac{\widehat{C}}{2}. When ABCD is cyclic, we have then \widehat{DGE}=\widehat{FAD}=\tfrac{\widehat{C}}{2} \Longrightarrow ADGE is cyclic, i.e. AD is antiparallel to EG WRT FE,FG \Longrightarrow perpendicular from F to AD goes through the circumcenter T of \triangle FGE. Similarly, perpendiculars from G,H,Eto DC,CB,BA pass through T.


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