diagram 1 is what the question has given me, and I've totalized it in diagram 2. Actually Diagram 2 is what i think must be the full diagram of this problem.
First of all the puzzle says: In the triangle $\triangle ABC$, bisector of smallest external angle, crosses the line &BC&(the largest line) in the point &D&. Now it says to find:$$\frac{S_\triangle ABD}{S_\triangle ABC}=?$$ So draw a complete diagram namely "diagram 2". Actually I draw a parallel line to $AB$, named $MD$. And also I streched the line $CA$ to $CM$. $M$ was named the cross point of the lines $DM$ and $CM$.
So now we have Thales's theorem in $\triangle CMD$.
I think that $\triangle ABC \thicksim \triangle AMD$. Because $\angle MAD=\angle DAB$.
And I Also know if we draw the height $AH$ in $\triangle ABC$ then: $$\frac{S_\triangle ABD}{S_\triangle ABC}=\frac{DB}{BC}$$ Now how to find $$DB=?$$
Your drawing is not perfect so you are missing some obvious things:
You are right, triangles $\triangle ABC$ and $\triangle MDC$ are similar because all their angles are equal. It means that:
$$\frac{MD}{AB}=\frac{MC}{AC}=\frac{MA+AC}{AC}\tag{1}$$
On the other side, triangle $\triangle MAD$ is isosceles because $\angle MDA=\angle DAB$ (as angles with parallel legs) and $\angle DAB=\angle DAM$. Because of that: $MD=MA=x$. Put it all now in (1) and you get:
$$\frac{x}{3}=\frac{x+5}{5}$$
$$x=\frac72$$
On the other side (Thales theorem):
$$\frac{DB}{BC}=\frac{MA}{AC}=\frac{\frac72}{5}=\frac7{10}=\frac{S_\triangle ABD}{S_\triangle ABC}$$