The measure of $\measuredangle B$ on the trapeze below

Question

For reference: In the ABCD trapeze(CB//AD), $m\measuredangle B = 4m \measuredangle D$ and $11AB + 5BC = 5AD$. Calculate $\measuredangle D$. (answer $26,5^\circ$}

My progress:

I could only find a relationship between the sides but I can't find a relationship with the angles

$Draw ~BE \parallel CD \implies \measuredangle BEA = \theta\\
Draw ~BF \measuredangle FBE = \theta \therefore \measuredangle ABF = 2\theta ~and~\measuredangle AFB = 2\theta\\
\triangle ABF~\triangle EFB \text{ are isosceles}\\
11m + 5a = 5(m+n+a)\rightarrow 11m+5a = 5m+5n+5a\rightarrow\\
11m = 5m + 5n \therefore m = \frac{5n}{6}$

Answer 1

Let $K\in AD$ such that $ABCK$ be a parallelogram.

Thus, by law of sines for $\Delta KCD$ we obtain: $$\frac{m}{\sin\theta}=\frac{\frac{11m}{5}}{\sin3\theta}.$$ Can you end it now?

I got $\theta=\arcsin\frac{1}{\sqrt5}.$