# Question #f1c54

Jun 27, 2017

See the proof below

#### Explanation:

If $P$ is the centroid of the triangle $A B C$, then

$A M = M C$, $\implies$, $\left[A M P\right] = \left[C P M\right]$

$C L = L B$, $\implies$, $\left[C P L\right] = \left[B P L\right]$

$A N = N B$, $\implies$, $\left[A P N\right] = \left[B P N\right]$

Also, we have

$\left[A C L\right] = \left[A L B\right]$, $\implies$

$\left[A M P\right] + \left[C P M\right] + \left[C P L\right] = \left[A P N\right] + \left[B P N\right] + \left[L P B\right]$

$2 \left[A M P\right] + \cancel{C P L} = 2 \left[B P N\right] + \cancel{L P B}$

$\left[A M P\right] = \left[B P N\right]$

$\left[C P M\right] = \left[A P N\right]$

Similarly,

$\left[A C N\right] = \left[B C N\right]$, $\implies$

$\left[A M P\right] + \left[C P M\right] + \left[A P N\right] = \left[C P L\right] + \left[B P N\right] + \left[B P L\right]$

$2 \left[C P M\right] + \cancel{A P N} = 2 \left[B P L\right] + \cancel{B P N}$

$\left[C M P\right] = \left[B P L\right]$

$\left[C P M\right] = \left[B P L\right]$

Finally,

$\left[A P N\right] = \left[B P L\right] = \left[C P M\right]$

$Q E D$