jmc

We know two of the angles in $△BCD$: $$m\angle CBD=80^{\circ },m\angle BCD=50^{\circ }.$$ Since the sum of angles in a triangle is $180^{\circ }$, we conclude that $m\angle BDC=180^{\circ }-(50^{\circ }+80^{\circ })=50^{\circ }$.

Therefore, $△BCD$ is isosceles with equal angles at $C$ and $D$, which implies that the sides opposite those angles ($BD$ and $BC$) are equal.

Since we are given $AB=BC$, we now know that $AB=BD$, which means that $△ABD$ is isosceles with equal angles at $A$ and $D$. Let $x=m\angle A$. Then the sum of angles in $△ABD$ is $180^{\circ }$, so $$x+x+30^{\circ }=180^{\circ }.$$ We can solve this equation to get $x=75^{\circ }$. So, $m\angle A=\boxed{75^{\circ }}$.