Quadrilateral $ABCD$ has $∠BDA = ∠CDB = 50^\circ, ∠DAC = 20^\circ$ and $∠CAB = 80^\circ.$ Find angles $∠BCA$ and $∠DBC.$ 
Quadrilateral $ABCD$ has $∠BDA=∠CDB=50°$, $∠DAC=20°$ and $∠CAB=80°$. Find angles $∠BCA$ and $∠DBC$.

This seems to be a very easy problem, but I am not able to solve it.
I am attaching a diagram which also contains all the simple angle chasing I have done.

I can not see any cyclic quads either or any parellelograms so I am not sure what to do.
 A: In a triangle, that is $\triangle ADC$, bisector of an interior angle, that is $BD$ bisector of $\angle ADC$, and bisector of non-adjacent two exterior angles, they are  $AB$ bisector of $\angle CAB$ and $CB$ bisector of $\angle ACB$ are concurrent.
Hence
$$\angle BCA=60^ {\circ}$$
$$\angle CBD=10^{\circ}$$

Given that point $B$ is the intersection of $DB$ and $AB$
$$\angle ADB = \angle CDB=50^{\circ}$$
$$\angle CAB = \angle BAQ=80^{\circ}$$
where $Q$ is a point on $DA$
Hence point $B$ is the centre of excircle of $\triangle ADC$ then $CB$ is the exterior bisector.
$$\angle ACB= \angle BCP = 60^{\circ}$$
where point P is on $DC$
A: 
While $ABCD$ is not a cyclic quadrilateral, if we extend $DC$ to point $E$ such that $\angle ABE = 80^0, \, ADEB$ is a cyclic quadrilateral. We then have,
$\angle AEB = \angle ADB = 50^0\,$ and $\angle BAE = \angle BDE = 50^0$.
So, $\triangle ABE$ is isosceles.
As $\angle BAE = 50^0, \angle DAE = 50^0$ so $\angle CAE = 30^0$.
Also, $\angle ACE = \angle ADE + \angle CAD = 120^0$
So, $\angle AED = 30^0 = \angle CAE$.
$\triangle ACE$ is isosceles.
So perpendicular from $B$ and $C$ will both meet $AE$ at its midpoint which means $BC$ is perpendicular to $AE$.
That leads to $\angle CBE = 180^0 - (90^0 + 50^0) = 40^0 \implies \angle DBC = 10^0$.
So we also have in $\triangle ABC, \angle ACB = 60^0$
