1.

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).

Answer»

SOLUTION :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.


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