Triangle $ABC$ is an isosceles right triangle with a right angle at $A$. Segments $BD$ and $BE$ trisect angle $ABC$. What is the degree measure of angle $BDE$?

[asy]draw((0,0)--(-1,0)--(-1,1)--(0,0)--cycle,linewidth(2));
draw((0,0)--dir(165)*1/cos(pi/12),linewidth(2));
draw((0,0)--dir(150)*1/cos(pi/6),linewidth(2));
draw((-1,.07)--(-.93,.07)--(-.93,0),linewidth(1));

label("A",(-1,0),SW);
label("B",(0,0),SE);
label("C",(-1,1),NW);
label("D",dir(150)*1/cos(pi/6),W);
label("E",dir(165)*1/cos(pi/12),W);
[/asy]
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 $\triangle ABD$ is a right triangle, $\angle ADB = 90-\angle ABD = 60^\circ$. Then, we see that $\angle ADB = \angle BDE,$ so we have $\angle BDE = \boxed{60^\circ}.$