I think I have a correct proof now.
Proof by contradiction: Assume to the contrary that two lines parallel to the same line are not parallel to each other. Without loss of generality, assume line m and line n are parallel to a line l, but m and n are not parallel to each other. Then, m and n intersect at a point, P that is not on line l. However, this contradicts Axiom 5 because two lines would be containing P and be parallel to l. So the assumption that m and n are not parallel was incorrect. Thus, m and n are parallel to l and also parallel to each other.
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
Best Answer
Given: $\triangle ABC$ with $CE \perp AB$, $\angle ACE = 20^\circ$ and $\angle BCE = 10^\circ$. Also, there is a point $D$ on $CE$ extend such that $\angle CDA = 50^\circ$. The question asks us to find $\angle ABD$.
Here is a construction that works.
We choose a point $G$ on $CI$ extend such that $\angle ABG = 10^\circ$. Then $ACBG$ is cyclic and a quick angle chasing shows that $\angle CAG = \angle CBG = 90^\circ$. So $CG$ must be a diameter of the circumcircle of $ACBG$ and its midpoint $O$ is the circumcenter. That leads to $\triangle AOB$ being an equilateral triangle.
As $\angle ADC = 50^\circ$ and $\angle DCF = 10^\circ$, we see that $\angle AFB = \angle BAF = 40^\circ$. So $BF = AB = BO$.
Since $\angle OBC = 20^\circ$, we have $\angle OFB = \angle BOF = 10^\circ$
Then notice that $\triangle AOC \cong \triangle OBF$ (by S-A-S)
So it follows that $AC = OF = CH$ and since $\angle GOF = \angle GCH = 30^\circ$, $OFHC$ is a parallelogram and we obtain that $\angle BFH = \angle BCO = 20^\circ$.
$\angle AHD = \angle DAH = 40^\circ$. As segment $BD$ subtends $40^\circ$ at point $H$ and $F$ on the same side of the segment, $BDFH$ is cyclic. That leads to $\angle BDH = \angle BFH = 20^\circ$.
Finally, $\angle ABD = \angle AHD + \angle BDH = 60^\circ$.