Ratios of a triangle question Consider a triangle $ABC$. Take a point $C′$ on the side $AB$. Assume that
$AC′:C′B=2$. Let $O$ be the midpoint of the segment $CC′$. Let $B′$ be the intersection point of the side $AC$ and the line $BO$. Find $AB′:B′C$ and $BO:OB′$.
I am trying to solve this question and I am a bit stuck because the only additional information I could gain from this question was that $CO$ and $OC'$ will be split in the $1:1$ ratio but thats not really helping me solve the rest of the question. Is there something I am missing ?
 A: 
Construct $PC'$ parallel to $BB'$. Let $AC' = 2x$, $C'B = x$ and $C'O = OC = y$.
Now, $ \Delta AC'P \sim \Delta ABB'$ and $\Delta CB'O\sim \Delta CPC'$.
$$\frac{CP}{CB'} = \frac{CC'}{CO}$$$$\frac{CB'+B'P}{CB'} = 2$$$$\frac{CB'}{B'P} = 1$$
$$B'C = PB'$$
$$\frac{AB'}{AP} = \frac{AB}{AC'}$$$$\frac{AP + PB'}{AP} = \frac32$$
$$\frac{PB'}{AP} = \frac12$$$$AP = 2PB'$$
$$\frac{AB'}{B'C} = \frac{AP+PB'}{PB'} = \frac{AP}{PB'}+1 = 2+1$$
$$\color{green}{ AB':B'C = 3}$$

$$\frac{BB'}{PC'} = \frac{AB}{AC'} = \frac32$$
$$PC' = \frac23BB'$$
$$\frac{PC'}{OB'} = \frac{CC'}{CO} = 2$$$$PC' = 2OB'$$$$\frac23BB' = 2OB'$$
$$BB' = 3OB'$$$$BO+OB' = 3OB'$$$$BO = 2OB'$$
$$\color{green}{BO:OB' = 2}$$
A: Menelaus's theorem applied to $(\triangle AC'C ,\;\overline{B-O-B'})$ gives $\frac{AB}{BC'}\frac{C'O}{OC}\frac{CB'}{B'A}=1\implies \frac{3}{1}\frac{1}{1}\frac{CB'}{'A}=1\implies AB':B'C=3:1$.
Menelaus's theorem applied to $(\triangle AB'B ,\;\overline{C-O-C'})$ gives $\frac{AC}{CB'}\frac{B'O}{OB}\frac{BC'}{C'A}=1\implies \frac{4}{1}\frac{B'O}{OB}\frac{1}{2}=1\implies BO:OB'=2:1$.
A: Extending $\overline{AB}$ by one-third of its length (ie, $|C'B|$) to $D$, and adding a few parallels to $\overline{BB'}$, handles both parts of the exercise pretty nicely. Details are left to the reader.

A: This problem can be approached using vectors.

Let $\vec{AB}=\boldsymbol a$ and $\vec{AC}=\boldsymbol b$.
Say $\vec{AB'}=x\vec{AC}$.
Therefore,
$$\vec{AB}+\vec{BB'}=x\vec{AC}$$
$$\boldsymbol a+y\vec{BO}=x\boldsymbol b\tag1$$
Now, $$\begin{align}\vec{BO}&=\vec{BC'}+\vec{C'O}\\&=\frac13\vec{BA}+\frac12\vec{C'C}\\&=\frac13\vec{BA}+\frac12(\vec{C'A}+\vec{AC})\\&=-\frac13\boldsymbol{a}+\frac12(-\frac23\boldsymbol a+\boldsymbol b)\\&=-\frac23\boldsymbol a+\frac12 \boldsymbol b\end{align}$$
Plugging this back to $(1)$ and solving, we get, $$\underbrace{\left(1-\frac23y\right)}_{\lambda}\boldsymbol a+\underbrace{\left(\frac y2-x\right)}_{\mu}\boldsymbol b=0.$$
Since $\boldsymbol a\ne\boldsymbol b$ and $\boldsymbol a\nparallel \boldsymbol b$ $\implies\lambda=\mu=0$.
Therefore $x=\frac34$ and $y=\frac32$.
Hence,
$$AB'=\frac34AC\implies AB':B'C=3:1$$
$$BB'=\frac32BO\implies BO:OB'=2:1$$
