What is 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

Your work is good. Now examine $\triangle ABF$ with sides $m, m$, and $6m/5$. Drop a perpendicular from the vertex, to get $3m/5$ for the adjacent and $m$ for the hypotenuse, so that $\cos 2\theta = 3/5$. Solve for $\theta.$

Answer 2

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