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I am confused on how to find the answer for this problem. So far what I believe would apply is the triangle inequality but I'm not sure on how to use it.

The figure $ABC$ is a triangle so as $BDC$. It is known that the
length of $AB = 8$ inches and $angle BAD = 2,angle BCD$ and
$angle,DBC = 3 angle BCD$. Find the smallest whole number that $BD$
can attain.

The alternatives given in my book are:

$beginarrayll
1.&7,textrminches\
2.&4,textrminches\
3.&8,textrminches\
4.&5,textrminches\
5.&6,textrminches\
endarray$

So far the only thing I could come up with was to "guess" it might be $4$ since, $angle BDA = 4omega$ could it be that since $BD$ is opposing $2omega$ hence that side might be $4$ inches. But that's how far I went. Needless to say that I'm unsure if that's a right way to go. Can somebody help me with this question?. Please try to include some drawing or sketch of the situation as I'd like to know what can I do to solve this problem relying mostly in geometry?.

A possible path without trigonometry.

EDIT Added explicitely the angles in the diagram, as required in comments.

1. Draw line $BE$ so that $angle EBC = angle ACB $.


2. 
   $triangle EBC$ is isoceles (why?); $triangle BDE$ is isosceles (why?); $triangle ABE$ is isosceles (why?)

Now use triangular inequality on $triangle BDE$, to get $2BD > 8$.

EDIT Since $triangle ABE$ is isosceles, $angle AEB cong angle EAB < 90°$, whereas one of the two between $angle BDA$ and $angle BDE$ is greater than or equal to $90°$. Since in any triangle the side opposite to the greater angle is greater, we can state either that $BD < BE$ or that $BD < AB$. So $BD< 8$.

EDIT (Thanks to Lubin for pointing this out in comment)

Once the fact that $4<BD<8$ has been proven, one should also check whether an integer solution between $4$ and $8$ does exist for $BD$. I will try to show, then, that in fact $BD = 5$ is a valid solution. Consider the Figure below.

Start at point $D$ and draw a circle $gamma_1$ centered in $D$ and with radius $5$; draw then a diameter of $gamma_1$, and let $E$ be one of its endpoints. Draw a circle $gamma_2$ centered in $E$ having radius $8$, which intersects $gamma_1$ in $B$, and the line $DE$ (at the opposite side of $D$) in $C$. Finally draw circle $gamma_3$ of radius $8$, centered in $B$. This intersects line $DE$ in two points. One is $E$; call the other one $A$. The triangle $triangle ABC$ satisfies the requirements, with $BD = 5$, as desired.

Since $angle ADB=4omega$, Law of Sines tells you that the length of $BD$ is $frac8sin(2omega)sin(4omega)$. Then, since $sin(2alpha)<2sinalpha$ for all acute angles $alpha$, the fraction is always greater than $4$, whereas for $omega=29^circ$, $angle ADB=116^circ$, and Law of Sines gives the length of $BD$ to be about $7.54$in. This length clearly varies continuously with $omega$ in the range $0<omega<30^circ$, so there are integer values for the length of $5,6,7$in. Your choice then is #4, $5$ inches.