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

Question

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

Answer 1

(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/AD+BP/BE+CP/CF
=(AD-PD)/AD+(BE-PE)/BE+(CF-PF)/CF
=1-PD/AD+1-PE/BE+1-PF/CF
=3-(PD/AD+PE/BE+PF/CF)
=3-1
=2