Let P be an interior point of a triangle ABC and AP, BP, CP meet the sides BC, CA, AB in D, E, F, respectively. Show that `(AP)/(PD)=(AF)/(FB)+(AE)/(EC)`.

Explanation

Since A, B, C, P are co-planar, there exists four scalars `x,y,z,w` not all zero simultaneously such that
`" "xveca+yvecb+zvecc+wvecp=0`
where `" " x+y+z+w=0`
Also, `" "(xveca+wvecp)/(x+w)=(yvecb+zvecc)/(y+z)`
Hence, `" "(AP)/(PD)=-(w)/(x)-1`
Also `" "(xveca +yvecb)/(x+y)=(zvecc+wvecp)/(z+w)`
`rArr" "(AF)/(FB)=(y)/(x)`
Similarly, `" "(AE)/(EC)=(z)/(x)`
Thus, to show that `-(w)/(x)-1=(y)/(x)+(z)/(x)`
`rArr" "x+y+z+w=0` which is true.
Hence proved.