In $\triangle ABC$, what is the value of $x + y$? [asy]
import olympiad;
size(7cm);

pair a = dir(76);
pair b = (-1, 0);
pair c = (1, 0);
pair o = (0, 0);

draw(a--b--c--cycle);
draw(a--o);

label("$A$", a, N); label("$B$", b, SW); label("$C$", c, SE);
label("$104^\circ$", o, 1.8 * NW + 0.4 * E);
label("$x^\circ$", b, 3 * E + NE + NE);
label("$y^\circ$", c, 2 * W + 2 * NW);

add(pathticks(b--o, s=3));
add(pathticks(c--o, s=3));
add(pathticks(a--o, s=3));

[/asy]
Since $\triangle BDA$ is isosceles, $\angle BAD = \angle ABD = x^\circ$.
Since $\triangle CDA$ is isosceles, $\angle CAD = \angle ACD = y^\circ$. [asy]
import olympiad;
size(7cm);

pair a = dir(76);
pair b = (-1, 0);
pair c = (1, 0);
pair o = (0, 0);

draw(a--b--c--cycle);
draw(a--o);

label("$A$", a, N); label("$B$", b, SW); label("$C$", c, SE); label("$D$", o, S);
label("$104^\circ$", o, 1.8 * NW + 0.4 * E);
label("$x^\circ$", b, 3 * E + NE + NE);
label("$y^\circ$", c, 2 * W + 2 * NW);

add(pathticks(b--o, s=3));
add(pathticks(c--o, s=3));
add(pathticks(a--o, s=3));

label("$x^\circ$", a, 3 * S + 2 * SW + W); label("$y^\circ$", a, 3 * S + SE);

[/asy] Therefore, $\angle BAC = (x + y)^\circ$.

Since the sum of the angles in $\triangle ABC$ is $180^\circ$, we have \begin{align*}
x + y + (x + y) &= 180\\
2x + 2y &= 180\\
x + y &= 90.
\end{align*}Therefore, $x + y = \boxed{90}$.