jmc

Since $AD=DC$, the angles in $△ADC$ opposite sides $AD$ and $DC$ are equal. Therefore, each of these angles is $22.5^{\circ }$, and $\angle ADC=(180-2\cdot 22.5{)}^{\circ }=135^{\circ }$.

Angles $\angle ADB$ and $\angle ADC$ add up to a straight angle, so $\angle ADB=45^{\circ }$.

Finally, since $BA=AD$, we have $\angle ABD=\angle ADB=\boxed{45^{\circ }}$.