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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). |
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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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