This problem is a particular case of a family of problems with broadly the same solution, so I will post this more general solution and then discuss particular instances of it.
Problem. $\angle BAC=3\angle CAD$; $\angle CBD=30^\circ$; $AB=AD$. What is $\angle DCA$?
Solution. Let $\alpha=\angle CAD$. $\triangle BDA$ is isosceles on base $BD$. Therefore $\angle DBA=\angle ADB=90^\circ-2\alpha$ and $\angle CBA=120^\circ-2\alpha$.
Let $E$ be on $BC$ such that $AE=AB$. Then $\triangle BEA$ is isosceles on base $BE$. Therefore $\angle AEB=\angle EBA=120^\circ-2\alpha$, so $\angle BAE=4\alpha-60^\circ$, so $\angle EAD=60^\circ$.
Therefore $\triangle AED$ is equilateral, so $\angle EAC=60^\circ-\alpha=\angle ACE$, so $\triangle CAE$ is isosceles on base $CA$, i.e. $CE=AE=DE$, so $\triangle CDE$ is isosceles on base $CD$. $\angle CED=2\alpha$, so $\angle DCE=90^\circ-\alpha$, so $\angle DCA=30^\circ$, which solves the problem. Note that $\angle DCA$ is independent of $\alpha$.
To adapt this to the current problem, relabel from $ABCD$ to $BCDA$ and specify $\alpha=19^\circ$.
If $\alpha$ is specified as $20^\circ$, and $\angle DBA$ as $50^\circ$, then the problem is [Langley]. $AB=AD$ is easily seen, and the proof proceeds as above. The above proof, but with angles as specified in Langley's problem, is due to J. W. Mercer.
If $\alpha$ is specified as $16^\circ$, then the problem is that at gogeometry. The point-lettering is the same, but the diagram is flipped.
[Langley] Langley, E. M. "Problem 644." Mathematical Gazette, 11: 173, 1922, according to David Darling
Let $\omega$, $O$ be the circumcircle and circumcenter of $\triangle ABC$, respectively. Let $P,Q,R,S$ be four points on the shorter arc $AC$ of $\omega$ dividing this arc into five equal parts.
First, we shall prove that $\triangle RSD$ is equilateral. Let $D'$ be a point inside $\omega$ such that $\triangle RSD'$ is equilateral. Also, let $E$ be inside $\omega$ such that $\triangle PQE$ is equilateral. Invoking symmetries we see that $\triangle D'SC \equiv \triangle D'RQ \equiv \triangle EQR \equiv \triangle EPA$. Note that $\angle EQR = \angle QRD'=\angle QRS-60^\circ = 168^\circ - 60^\circ = 108^\circ$. Hence $\angle D'QR = 90^\circ - \frac 12\angle QRD' = 36^\circ$ and $\angle EQD'=108^\circ - 36^\circ = 72^\circ$. But also $\angle D'EQ = 180^\circ - \angle EQR = 180^\circ - 108^\circ = 72^\circ$. Hence $ED'Q$ is isosceles with $QD'=ED'$. Again, using symmetries we see that $AED'C$ is an isosceles trapezoid with $AE=ED'=D'C$. We have $\angle ACD'=\angle SCD' - \angle SCA = 36^\circ - 24^\circ = 12^\circ$. Since $AED'C$ is an isosceles trapezoid, it is cyclic and since $AE=ED'=D'C$, it follows that $\angle D'AC = \frac 12 \angle EAC = \frac 12 \angle ACD'=6^\circ$. Hence $D'$ coincides with $D$.
Now comes my favourite part. Some angle chasing shows that $\angle QCE = 18^\circ = \angle DCB$ and $\angle DQC = 24^\circ = \angle BQE$. Hence $D$ and $E$ are isogonal conjugates in $\triangle BQC$. It follows that $\angle CBD = \angle EBQ$.
Choose $T$ on $\omega$ so that $BT$ is a diameter. Clearly, $\triangle BQE$ is symmetric to $\triangle TRD$ with respect to perpendicular bisector of $QR$. In particular, $\angle RTD = \angle EBQ$.
Let $RT$ intersect $BC$ at $X$. Since $\angle CBD = \angle EBQ = \angle RTD$, quadrilateral $BDXT$ is cyclic. Hence $\angle BDT = \angle BXT$. Then some angle chasing shows that $\angle DOB = 102^\circ = \angle BXT = \angle BDT$. This precisely means that the circumcircle of $DOT$ is tangent to $BD$ at $D$. Tangent-secant theorem yields $BD^2=BO\cdot BT = BO \cdot 2BO = 2BO^2$. Hence $$\frac{BD}{AB} = \frac{BD}{BO} = \sqrt 2,$$
as desired.
Best Answer
The solutions are not as trivial as one would expect from the statement. It's called Langley's problem of adventitious angles first posed in The Mathematical Gazette in 1922.
Check out An Intriguing Geometry Problem by Tom Rike.