jmc

Let $x$ represent the length of each of the edges. If we draw the diagonal $\overline{BD}$, we create the right triangle $BCD$. Since $\overline{BC}\cong \overline{DC}$, $△BCD$ is a 45-45-90 right triangle, which means the hypotenuse has a length of $\sqrt{2}$ times the length of each leg. So $BD=x\sqrt{2}$. Since $△ABD$ is an isosceles triangle, we know that drawing a segment from $A$ to the midpoint of $\overline{BD}$ splits the triangles into two congruent right triangles. Each right triangle has a hypotenuse of length $x$ and a leg of length $\frac{BD}{2}=\frac{x\sqrt{2}}{2}=\frac{x}{\sqrt{2}}$. Notice that the hypotenuse has a length of $\sqrt{2}$ times the length of the bottom leg, which means the triangle is another 45-45-90 right triangle. That means the degree measure of angle $ABD$ is $\boxed{45^{\circ }}$.