From the figure, we know that $\displaystyle \angle ABC = \angle ACB = 50^o$
So, $\displaystyle \angle DCB = 30^o$
From triangle ACD, we know that $\displaystyle \angle ADC = 150^o$
So, we get some equations now.
Let's give some names to the other angles:
$\displaystyle \angle ABD = a$
$\displaystyle \angle DBC = b$
$\displaystyle \angle BDC = c$
From triangle ABD, we know that: $\displaystyle 70^o + \alpha + a = 180^o$
From triangle DBC, we know that: $\displaystyle b + c + 30^o = 180^o$
And lastly, we know that $\displaystyle \alpha + 150^o + c = 360^o$
and $\displaystyle b + c = 50^o$
That should be a + b = 50º.
There is enough information to find alpha