Think about two folds

Draw $EF$ parallel to $AD$ as shown.

It can be easily shown that $\Delta BOD$ and $\Delta BEF$ are similar triangles. Thus by the basic proportionality theorem, we have

$\begin{aligned} & \dfrac{BO}{BE} &=& \dfrac{OD}{EF} \\ \\ \implies & \dfrac 12 &=& \dfrac{OD}{EF} \\ \\ \implies & \color{#3D99F6}{EF} &=& \color{#3D99F6}{2 OD} \end{aligned}$

Also, $\Delta CEF$ and $\Delta CAD$ are similar. Thus by BPT,

$\begin{aligned} & \dfrac{CE}{CA} &=& \dfrac{EF}{AD} \\ \\ \implies & \dfrac 12 &=& \dfrac{EF}{AD} \\ \\ \implies & AD &=& 2 \color{#3D99F6}{EF} \\ \\ \implies & AO+OD &=& 4OD \\ \\ \implies & AO &=& 3OD \\ \\ \implies & \dfrac{AO}{OD} &=& \boxed{\dfrac 31} \end{aligned}$

Add $D,E$

Now let $\Delta$ $ABE=x$ and $\Delta$ $BDO=a$

So $\Delta ABE=\Delta BEC=x$

So $\Delta BDO=\Delta DEO=a$

Then $\Delta ADE=\Delta DEC=x+a$

So $\Delta BEC=x+3a=2x$

Then $a=x/3$

So $AO:OD=x:a=1:1/3=3:1$

Using Menelaus's Theorem in $\Delta \text{BEC}$ with transversal $\text{AD}$ ,

$\displaystyle \dfrac{CA}{EA} \times \dfrac{EO}{BO} \times \dfrac{BD}{DC}=1$

Using the given conditions this simplifies to $\text{DC}=2\text{BD}$ , and similarly for $\Delta \text{ACD}$ with transversal $\text{BE}$ we get,

$\displaystyle\begin{aligned} &\dfrac{CB}{BD}\times \dfrac{DO}{OA}\times \dfrac{AE}{EC} = 1 \\ &\dfrac{BD+DC}{BD} = \dfrac{AO}{OD} \\ & \dfrac{AO}{OD} = \dfrac{3BD}{BD}=\boxed{3} \end{aligned}$