In the given triangle, the following side-length ratios are known,
$$AE ∶ EB = 3 ∶ 4,\>\>\> BD ∶ DC = 5 ∶ 6,\>\>\> AF ∶ FC = 2 ∶ 3$$
Then find the ratio AO ∶ OD.
I can solve it by mass point geometry but i want to know if there is any other method available. MpG feels like finding center of mass in physics, unlike rigorous mathematics.
On
Here's an alternative method using similar triangles, which may be a more familiar concept to the OP. In the triangle $ABC$, draw lines $BG$ and $DH$ parallel to $EF$, to intersect $AC$ at $G$ and $H$ respectively, as shown in the diagram below.
We have $\ AF = \frac{2}{5}AC\ $, and therefore $$ FG = \frac{4}{3}AF = \frac{8}{15}AC $$ by the similarity of triangles $AEF$ and $ABG$. Then $GC= AC-AG$$=\frac{1}{15}AC$, and $$ GH = \frac{5}{11}GC = \frac{1}{33}AC $$ from the similarity of triangles $CBG$ and $CDH$. Now from the similarity of triangles $AOF$ and $ADH$ we get $$ AO:OD = AF:FH = \frac{2}{5}:\left(\frac{8}{15}+\frac{1}{33}\right)=22:31\ . $$
On
I’ll provide a simple solution involving the Menelaus’ Theorem (with non-signed lengths): 
Extend $EF$ to meet $CB$ extended at M.
Given $\dfrac {BD}{DC}=\dfrac56$, let $BD=5X, DC=6X$ and let $MB=uX$. Then by Menelaus in $\triangle ABC$, $$\frac{MC}{MB}\cdot \frac{BE}{EA}\cdot \frac{AF}{FC}=1$$ so that $$\frac{u+5+6}{u}\cdot \frac43\cdot \frac23=1$$ so we find that $u=88$.
Again by Menelaus in $\triangle ABD$, $$\frac{MD}{MB}\cdot \frac{BE}{EA}\cdot \frac{AO}{OD}=1$$ so
$$\frac{u+5}{u}\cdot \frac43\cdot \frac{AO}{OD}=1$$ or $$\frac{93}{88}\cdot \frac43\cdot \frac{AO}{OD}=1$$ or $$\frac{AO}{OD}=\frac{22}{31}.$$
Denote various areas as $[\cdot]$ and use area ratios below to evaluate
\begin{align} \frac{AO}{OD} =& \ \frac{[AEF]}{[EDF]} = \frac{[AEF]}{[EBCF]-[FDC]-[EBD]}\\ = &\ \frac{\frac 25 \frac 37 [ABC]} {\left( 1-\frac 25 \frac 37 \right)[ABC] - \frac 35 \frac{6}{11}[ABC] - \frac 47 \frac{5}{11}[ABC]} \\ = &\ \frac{\frac{6}{35}} {\frac{25}{35} - \frac{18}{55} - \frac{20}{77}}=\frac{22}{31} \end{align}