Các điểm http://dientuvietnam.net/cgi-bin/mimetex.cgi?\large\dfrac{AD}{DB}.\dfrac{AE}{EB}=(\dfrac{AC}{CB})^2
Chứng minh rằng: http://dientuvietnam.net/cgi-bin/mimetex.cgi?\large\widehat{ACD}=\widehat{BCE}
Poland 1998
Bắt đầu bởi Merlyn, 07-05-2006 - 11:14
#1
Đã gửi 07-05-2006 - 11:14
#2
Đã gửi 08-05-2006 - 14:07
Giả sử http://dientuvietnam.net/cgi-bin/mimetex.cgi?D\in{[AE]}
Ta có :http://dientuvietnam.net/cgi-bin/mimetex.cgi?\dfrac{dt(ACD)}{dt(BCD)}=\dfrac{DA}{DB}=\dfrac{CA.Sin(\alpha)}{CB.Sin(\beta+x)}
Và: http://dientuvietnam.net/cgi-bin/mimetex.cgi?\dfrac{dt(ACE)}{dt(BCE)}=\dfrac{EA}{EB}=\dfrac{CA.Sin(\alpha+x)}{CB.Sin(\beta)}
(TRong đó:http://dientuvietnam.net/cgi-bin/mimetex.cgi?\alpha=\widehat{ACD};x=\widehat{DCE};\beta=\widehat{BCE})
TỪ ĐÓ:http://dientuvietnam.net/cgi-bin/mimetex.cgi?Sin(\alpha).Sin(\alpha+x)=Sin(\beta).Sin(\beta+x)
Suy ra:http://dientuvietnam.net/cgi-bin/mimetex.cgi?Cos(2\alpha+x)=Cos(2\beta+x) Nên:http://dientuvietnam.net/cgi-bin/mimetex.cgi?\alpha=\beta
Ta có :http://dientuvietnam.net/cgi-bin/mimetex.cgi?\dfrac{dt(ACD)}{dt(BCD)}=\dfrac{DA}{DB}=\dfrac{CA.Sin(\alpha)}{CB.Sin(\beta+x)}
Và: http://dientuvietnam.net/cgi-bin/mimetex.cgi?\dfrac{dt(ACE)}{dt(BCE)}=\dfrac{EA}{EB}=\dfrac{CA.Sin(\alpha+x)}{CB.Sin(\beta)}
(TRong đó:http://dientuvietnam.net/cgi-bin/mimetex.cgi?\alpha=\widehat{ACD};x=\widehat{DCE};\beta=\widehat{BCE})
TỪ ĐÓ:http://dientuvietnam.net/cgi-bin/mimetex.cgi?Sin(\alpha).Sin(\alpha+x)=Sin(\beta).Sin(\beta+x)
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