Cho P nằm trong tg ABC.AP,BP,CP cắt các cạnh t/ứng tại D,E,F.CMR:
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Bắt đầu bởi dhkhtn-tnt, 25-10-2005 - 17:52
#1
Đã gửi 25-10-2005 - 17:52
#2
Đã gửi 29-10-2005 - 14:28
Qua A vẽ dt//BC cắt DE DF ở M N ta có:
http://dientuvietnam.net/cgi-bin/mimetex.cgi?AM=\dfrac{EA}{EC}.DC
http://dientuvietnam.net/cgi-bin/mimetex.cgi?AN=\dfrac{FA}{FD}.DB
mặt khác theo định lý xeva có http://dientuvietnam.net/cgi-bin/mimetex.cgi?\dfrac{DB}{DC}.\dfrac{FA}{FB}.\dfrac{EC}{EA}=1
Suy ra AM=AN
http://dientuvietnam.net/cgi-bin/mimetex.cgi?\dfrac{1}{PD}=\dfrac{1}{AD}+\dfrac{1}{BD}+\dfrac{1}{CD}
hay http://dientuvietnam.net/cgi-bin/mimetex.cgi?\dfrac{EA}{EC}=\dfrac{DA}{DC}
DE là fg góc http://dientuvietnam.net/cgi-bin/mimetex.cgi?\widehat{ADC} hay http://dientuvietnam.net/cgi-bin/mimetex.cgi?AD=AM=AN\Leftrightarrow{\widehat{EDF}=90^{0}}
http://dientuvietnam.net/cgi-bin/mimetex.cgi?AM=\dfrac{EA}{EC}.DC
http://dientuvietnam.net/cgi-bin/mimetex.cgi?AN=\dfrac{FA}{FD}.DB
mặt khác theo định lý xeva có http://dientuvietnam.net/cgi-bin/mimetex.cgi?\dfrac{DB}{DC}.\dfrac{FA}{FB}.\dfrac{EC}{EA}=1
Suy ra AM=AN
http://dientuvietnam.net/cgi-bin/mimetex.cgi?\dfrac{1}{PD}=\dfrac{1}{AD}+\dfrac{1}{BD}+\dfrac{1}{CD}
hay http://dientuvietnam.net/cgi-bin/mimetex.cgi?\dfrac{EA}{EC}=\dfrac{DA}{DC}
DE là fg góc http://dientuvietnam.net/cgi-bin/mimetex.cgi?\widehat{ADC} hay http://dientuvietnam.net/cgi-bin/mimetex.cgi?AD=AM=AN\Leftrightarrow{\widehat{EDF}=90^{0}}
Qua A vẽ dt//BC cắt DE DF ở M N ta có: [TeX]AM=\dfrac{EA}{EC}.DC[/TeX] [TeX]AN=\dfrac{FA}{FD}.DB[/TeX] mặt khác theo định lý xeva có [TeX]\dfrac{DB}{DC}.\dfrac{FA}{FB}.\dfrac{EC}{EA}=1[/TeX] Suy ra AM=AN [TeX]\dfrac{1}{PD}=\dfrac{1}{AD}+\dfrac{1}{BD}+\dfrac{1}{CD}[/TeX] hay [TeX]\dfrac{EA}{EC}=\dfrac{DA}{DC}[/TeX] DE là fg góc [TeX]\widehat{ADC}[/TeX] hay [TeX]AD=AM=AN\Leftrightarrow{\widehat{EDF}=90^{0}}[/TeX]
Bài viết đã được chỉnh sửa nội dung bởi MrMATH: 29-10-2005 - 16:02
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