Lời giải:

Ta có: $\dfrac{S_{BMN}}{S_{ABC}}=\dfrac{\dfrac{1}{2}BM.BN.\sin B}{\dfrac{1}{2}BA.BC.\sin B}=\dfrac{BM}{BA}.\dfrac{BN}{BC}=\dfrac{2}{5}.\dfrac{2}{3}=\dfrac{4}{15}$

$\Rightarrow S_{BMN}=\dfrac{4}{15}S_{ABC}$ (1)

Ta có: $\dfrac{S_{BMK}}{S_{BAO}}=\dfrac{\dfrac{1}{2}BM.BK.\sin \widehat{MBK}}{\dfrac{1}{2}BA.BO.\sin\widehat{ABO}}=\dfrac{BM}{BA}.\dfrac{BK}{BO}=\dfrac{2}{5}\dfrac{BK}{BO}$

Tương tự: $\dfrac{S_{BKN}}{S_{BOC}}=\dfrac{BK}{BO}.\dfrac{BN}{BC}=\dfrac{2}{3}\dfrac{BK}{BO}$

Mà $S_{ABO}=S_{ACO}=\dfrac{1}{2}S_{ABC}$ $(=\dfrac{1}{2}d_{A,BO}.BO=\dfrac{1}{2}d_{C,BO}.BO$

$\Rightarrow S_{BMN}=S_{BMK}+S_{BKN}$

$=\dfrac{2}{5}\dfrac{BK}{BO}\dfrac{1}{2}S_{ABC}+\dfrac{2}{3}\dfrac{BK}{BO}\dfrac{1}{2}S_{ABC}$

$=\dfrac{8}{15}\dfrac{BK}{BO}S_{ABC}$ (2)

Từ (1) và (2) suy ra $\dfrac{BK}{BO}=\dfrac{1}{2}$

Ta có: $\dfrac{S_{JKD}}{S_{JKS}}=\dfrac{\dfrac{1}{2}d_{D,KJ}.KJ}{\dfrac{1}{2}d_{S,KJ}.KJ}=\dfrac{DP}{SP}=\dfrac{1}{5}$

Ta có: $\dfrac{S_{JKS}}{S_{JDS}}=\dfrac{\dfrac{1}{2}d_{(K,SJ)}.SJ}{\dfrac{1}{2}d_{(D,SJ)}.SJ}=\dfrac{KO}{DO}=\dfrac{1}{2}$

Ta có:

$S_{SKD}=S_{JKD}+S_{JKS}+S_{JDS}$

$=S_{JKD}+5S_{JKD}+2.5.S_{JKD}$

$=16S_{JKD} $

$\Rightarrow \dfrac{JO}{SO}=\dfrac{S_{JKD}}{S_{SKD}}=\dfrac{1}{16}$

$\Rightarrow \dfrac{SJ}{SO}=\dfrac{15}{16}$