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