Question #f1c54

1 Answer
Jun 27, 2017

Answer:

See the proof below

Explanation:

enter image source here

If #P# is the centroid of the triangle #ABC#, then

#AM=MC#, #=>#, #[AMP]=[CPM]#

#CL=LB#, #=>#, #[CPL]=[BPL]#

#AN=NB#, #=>#, #[APN]=[BPN]#

Also, we have

#[ACL]=[ALB]#, #=>#

#[AMP]+[CPM]+[CPL]=[APN]+[BPN]+[LPB]#

#2[AMP]+cancel[CPL]=2[BPN]+cancel[LPB]#

#[AMP]=[BPN]#

#[CPM]=[APN]#

Similarly,

#[ACN]=[BCN]#, #=>#

#[AMP]+[CPM]+[APN]=[CPL]+[BPN]+[BPL]#

#2[CPM]+cancel[APN]=2[BPL]+cancel[BPN]#

#[CMP]=[BPL]#

#[CPM]=[BPL]#

Finally,

#[APN]=[BPL]=[CPM]#

#QED#