Cfrgtkky

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):

$$frac{DB}{BC}=frac{MA}{AC}=frac{frac72}{5}=frac7{10}=frac{S_triangle ABD}{S_triangle ABC}$$