Triangles $ABC$ and $ABD$ are  isosceles with $AB=AC=BD$, and $\overline{BD}$ intersects $\overline{AC}$ at $E$. If $\overline{BD}\perp\overline{AC}$, then what is the value of $\angle C+\angle D$? [asy]
import olympiad; import geometry; size(150); defaultpen(linewidth(0.8));
draw(origin--(dir(70))--(2*Cos(70),0)--cycle);
draw(origin--(dir(20))--(dir(70))--cycle);
pair E = intersectionpoint(origin--dir(20),dir(70)--(2*Cos(70),0));
dot("$E$",E);
pair A = dir(70);
pair B = origin;
pair C = (2*Cos(70),0);
pair D = dir(20);
dot("$A$",A,N);
dot("$B$",B,SW);
dot("$C$",C,SE);
dot("$D$",D,E);
draw(rightanglemark(B,E,A,1.5));
[/asy]
Let $\angle ABD=x$ and $\angle BAC=y$. Since triangles $ABC$ and $ABD$ are isosceles, $\angle C=(180^\circ-y)/2$ and $\angle D=(180^\circ-x)/2$.  Then, noting that $x+y=90^\circ$, we have $$\angle C+\angle D=(360^\circ-(x+y))/2=\boxed{135^\circ}.$$ [asy] import olympiad; import geometry; size(150); defaultpen(linewidth(0.8)); draw(origin--(dir(70))--(2*Cos(70),0)--cycle); draw(origin--(dir(20))--(dir(70))--cycle); pair E = intersectionpoint(origin--dir(20),dir(70)--(2*Cos(70),0)); dot("$E$",E); pair A = dir(70); pair B = origin; pair C = (2*Cos(70),0); pair D = dir(20); dot("$A$",A,N); dot("$B$",B,SW); dot("$C$",C,SE); dot("$D$",D,E); draw(rightanglemark(B,E,A,1.5));
label("$x$",B+(0.06,0.07),NE);label("$y$",A-(0,.1),S); [/asy]