No, I could not. Most angles can be found through basic Geometry, but the last two (including the desired $\angle BDA$) will require Trigonometry. I did try a system of equations based on $(1)$ and $(4)$, but it got to $68^\circ=B\hat{D}A+68^\circ-B\hat{D}A\iff0=0$: nothing useful.
Let's name the point of the crossing diagonals $O$.
$$
C\hat{O}B=180^\circ-B\hat{C}A-D\hat{B}C=112^\circ\ \ \ \ \ \ \ \ \ \ \ \ \ (1)\\
B\hat{O}A=180^\circ-C\hat{O}B=68^\circ\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)\\
D\hat{O}C=B\hat{O}A=68^\circ\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (3)\\
A\hat{O}D=C\hat{O}B=112^\circ\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (3)\\
C\hat{D}B=180^\circ-A\hat{C}D-D\hat{O}C=64^\circ\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)\\
B\hat{A}C=180^\circ-B\hat{O}A-A\hat{B}D=74^\circ\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)\\
$$
(the numbers between parentheses are the basic Geometry rules used - referred below. I use the hat to refer to the vertice of the angle, and the 3-letter combination to refer to the amplitude of the respective angle, being the letters' order always counterclockwise)
$(1)$ the sum of the internal angles of any triangle is $180^\circ$
$(2)$ supplementary angles sum to $180^\circ$
$(3)$ vertically opposite angles are equal in amplitude
$(4)$ the sum of the internal angles of any quadrilateral is $360^\circ$
By $(1)$ (and $(4)$ works too) we know $68^\circ=D\hat{A}C+B\hat{D}A$, however, neither algebra nor basic Geometry take us any further (I thought there was a rule in Geometry for this, however, it requires the existence of some parallel lines, like the ones in a trapezoid).
The way harder method you used requires Trigonometry, so you certainly know all the rest (I can add it, but you didn't request it, and I'm a little rusty in that area).
Best Answer
The best proof by far:
$B$ is clearly the $D$ excenter of $\triangle ADC$, therefore $DB$ is an angle bissector of that triangle.
So: $\angle CDB = \frac{180^{\circ} - 60^{\circ} - 40^{\circ}}2 = 40^{\circ}$
$\angle DBC = 180^{\circ} - 40^{\circ} - 120^{\circ} = 20^{\circ}$