I am trying to find out $BD$ when in given triangle $DE =12$ ,$EC=11$,$\angle AED=75$ and $\angle ABC =70$. the given side DE is not parallel to the base BC ,so the ratio of $\frac{AD}{DB} \ne \frac{AE}{EC}$
I am thinking about making DE parallel to BC by rotating the line by 5 degrees and the new point which will be present on AC will be DM where $EM=\frac{5}{360}DE$ so now the line will become parallel to BC.
$\frac {\sin x}{12} = \frac {\sin 75}{n} = \frac {\sin (105-x)}m$
and $\frac {\sin x}{p} = \frac {\sin 75}{11 + m} = \frac {\sin (110-x)}{K + n}$
$m^2 + n^2 = 12^2 + 2nm\cos x$
$m^2 + 12^2 = n^2 + 24m\cos 75$
$n^2 + 12^2 = m^2 + 24n\cos (105-x)$
$(n+K)^2 + (m+11)^2 = p^2 + 2(n+K)(m+11)\cos x$
$(n+K)^2 + p^2 = (m+11)^2 + 2(n+K)p\cos 70$
$(m+11)^2 + p^2 = (n+K)^2 + 2(m+11)p\cos (110 - x)$
Now it's... work.
Or using your idea in the comments, use the second image.