$\angle ABD=38°, \angle DBC=46°, \angle BCA=22°, \angle ACD=48°,$ then find $\angle BDA$

I'm interested in Langley's problem.

I've been struggling to solve the following question and I've just got the result by a tedious calculation using Euler's formula $e^{i\pi x/{180}}=\cos{x°}+\sin{x°}$.

In a convex quadrilateral $ABCD$, letting $\angle ABD=38°, \angle DBC=46°, \angle BCA=22°, \angle ACD=48°,$ then find $\angle BDA$.

The answer I got is $\angle BDA=18°$. Then, here is my question.

Question: Could you show me how to prove this by the way of elementary geometry?

I've tried, but it seems very difficult.

Solution 1:

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).