jmc

From the given information, we see that $\angle ACB=\angle ABC=45^{\circ }$. Thus $\angle ABE=\angle EBD=\angle DBC=15^{\circ }$. So we obtain that $\angle ABD=30^{\circ }$, and since $△ABD$ is a right triangle, $\angle ADB=90-\angle ABD=60^{\circ }$. Then, we see that $\angle ADB=\angle BDE,$ so we have $\angle BDE=\boxed{60^{\circ }}.$