如图,P为△ABC内任一点,过P作AD,BE,CF分别与BC,AC,AB交于点D,E,F.求证:PD/AD+PE/BE+PF/CF=1

问题描述:

如图,P为△ABC内任一点,过P作AD,BE,CF分别与BC,AC,AB交于点D,E,F.求证:PD/AD+PE/BE+PF/CF=1
(2)AP/AP+BP/BE+CP/CF=2

(1)过P作MN∥BC,分别交AB,AC于M,N则△EPN∽△EBC,△FMP∽△FBC,∴PE/BE=PN/BC,PF/CF=PM/BC,∴PE/BE+PF/CF=MN/BC又∵△AMN∽△ABC,∴AM/AB=MN/BC又∵PD/AD=BM/AB,∴PD/AD+PE/BE+PF/CF=BM/AB+AM/AB=AB/AB=1 (2)AP...