Question #f1c54

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Answer 1 · 2017-06-27T16:22:00

See the proof below

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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,

$[APN]=[BPL]=[CPM]$

$QED$