Let $\square ABCD$ be a convex quadrilateral with $\angle DAC = \angle ACD = 17^{\circ}$; $\angle CAB = 30^{\circ}$; and $\angle BCA = 43^{\circ}$. Compute $\angle ABD$.
What I have so far:
Since $\angle DAC = \angle ACD = 17^{\circ}$ , points $A$, $C$, and $D$ form an isosceles triangle. We can look at the circle with center $D$ and radius $|DA|=|DC|$.
I believe this can help because one can then look at the angles subtended by the arc $AC$ and use them to gain further information. (I have seen this done in a similar problem.) I can't seem to find any angles that are $\frac{146}2=73^{\circ}$, though, so this is probably wrong.
Any help is appreciated
reflex $\angle ADC=360^{\circ}-146^{\circ}=214^{\circ}=2\times 107^{\circ}=2\times \angle ABC$
Since the angle subtended at the center of a circle is double the angle subtended at the circumference, above proves that our constructed circle also passes through the point $B$. So, $|DA|=|DB|=|DC|$ which in turn means that $\angle ABD=\angle BAD=\angle DAC+\angle BAC=47^{\circ}$.