Đáp án: $ S_{APN}=\dfrac1{15}S_{ABC}$

Giải thích các bước giải:

Ta có:

$BM=MC$

$\to S_{PBM}=S_{PCM}$

$\to$Khoảng cách từ $B, C$ đến $PM$ bằng nhau

$\to S_{PBN}=S_{PCN}$

Vì $AN=\dfrac15AC$

$\to S_{ANB}=\dfrac15S_{ABC}$

$\to$Khoảng cách từ $N$ đến $AB=\dfrac15$ khoảng cách từ $C$ đến $AB$

$\to S_{NPB}=\dfrac15S_{PBC}$

$\to 5S_{NPB}=S_{PBC}$

$\to 5S_{NPB}=S_{PNB}+S_{PNC}+S_{NBC}$

$\to 5S_{NPB}=2S_{PNB}+S_{NBC}$

$\to 3S_{NPB}=S_{NBC}$

Ta có: $AN=\dfrac15AC$

$\to AC-AN=AC-\dfrac15AC$

$\to NC=\dfrac45AC$

$\to S_{NBC}=\dfrac45S_{ABC}$

$\to 3S_{NBP}=\dfrac45S_{ABC}$

$\to 3S_{NBP}=\dfrac4{5}S_{ABC}$

$\to 3(S_{APN}+S_{ANB})=\dfrac4{5}S_{ABC}$

$\to 3(S_{APN}+\dfrac15S_{ABC})=\dfrac4{5}S_{ABC}$

$\to 3S_{APN}+\dfrac35S_{ABC}=\dfrac4{5}S_{ABC}$

$\to 3S_{APN}=\dfrac4{5}S_{ABC}-\dfrac35S_{ABC}$  

$\to 3S_{APN}=\dfrac15S_{ABC}$  

$\to S_{APN}=\dfrac1{15}S_{ABC}$