Finding the ratios of the segments in a triangle In triangle $ABC$ take $C'$ on $AB$ with $AC':C'B=2$. Let $O$ be the midpoint of $CC'$ and $B'$ be the intersection of $AC$ and $BO$. Find $AB':B'C$ and $BO:OB'$.
To solve this I tried to use a centre of mass technique, which gave me $AB':B'C=3:1$ but I’m not sure if that’s correct and I can’t really justify it, is there any other easier way to show this?

 A: Consider $M$ the midpoint of $C'A$. Then $O$ midpoint of $CC'$ and hence $MO\parallel AC$.
And clearly by Thales theorem $\dfrac{BO}{OB'}=\dfrac{BM}{MA}=\dfrac{2AB/3}{AB/3}=2$.
We consider the line through $C'$ parallel to $AC$. This intersects $BB'$ at $H$.
Then $HO=BO/2=OB'$ and hence the triangles $C'OH$ and $OB'C$ are equal.
Thus $B'C=C'H=MO/2=AC/4$ and hence $\dfrac{AB'}{B'C}=\dfrac{3AC/4}{AC/4}=3$.

A: Let us apply in the given situation the theorem of
Menelaus
for the triangle $\Delta AC'C$ w.r.t. the "secant" line $BOB'$, the fractions are fractions of oriented segments of the corresponding lines, so the sign is negative when the intersection point with the "secant" is inside the sides:
$$
1
=
\frac{BA}{BC'}\cdot
\frac{OC'}{OC}\cdot
\frac{B'C}{B'A}\cdot
=
\frac31\cdot
(-1)\cdot
\frac{B'C}{B'A}\cdot
\ .
$$
This gives $B'A:B'C = -3:1$, if unsigned segments are considered, then the result is $3:1$.
In a similar manner, to get $OB:OB'$ we use Menelaus in $\Delta ABB'$ with the "secant" $COC'$:
$$
1
=
\frac{CB'}{CA}\cdot
\frac{C'A}{C'B}\cdot
\frac{OB}{OB'}\cdot
=
\frac14\cdot
(-2)\cdot
\frac{OB}{OB'}\cdot
\ .
$$
So $OB:OB'=-2$. (Ignore the sign if unsigned segments are used.)

In this case, and in similar cases, Menelaus goes straightforward to the solution.
Here is also a picture illustrating this situation:

