A bridge is built by suspending a plank of wood between two triangular wedges with equal heights, as in the following diagram: [asy]
import olympiad;
import math;

// Draw triangles
pair A = (0, 1);
pair B = (-cos(1.3962), 0);
pair C = (cos(1.3962), 0);
pair D = (2, 1);
pair E = (2-cos(1.3089), 0);
pair F = (2+cos(1.3089), 0);
draw(A--B--C--cycle);
draw(D--E--F--cycle);
draw(A--D);
label('$A$',A,N);
label('$B$',B,S);
label('$C$',C,S);
label('$D$',D,N);
label('$E$',E,S);
label('$F$',F,S);
[/asy] If $AB = AC$ and $DE = DF,$ and we have $\angle BAC = 20^\circ$ and $\angle EDF = 30^\circ,$ then what is $\angle DAC + \angle ADE$?
There are several ways to proceed, and here is one. Since $\triangle ABC$ and $\triangle DEF$ are both isosceles, it should be easy to find that $\angle B = \angle C = 80^\circ$ and $\angle E = \angle F = 75^\circ.$ Now, connect $C$ and $E$:

[asy]
import olympiad;
import math;

// Draw triangles
pair A = (0, 1);
pair B = (-cos(1.3962), 0);
pair C = (cos(1.3962), 0);
pair D = (2, 1);
pair E = (2-cos(1.3089), 0);
pair F = (2+cos(1.3089), 0);
draw(A--B--C--cycle);
draw(D--E--F--cycle);
draw(A--D);
draw(C--E);
label('$A$',A,N);
label('$B$',B,S);
label('$C$',C,S);
label('$D$',D,N);
label('$E$',E,S);
label('$F$',F,S);
[/asy] Since the two triangular wedges have the same height, we see that $AD \parallel CE,$ thus $\angle DAC = \angle ACB = 80^\circ.$ Likewise, $\angle ADE = \angle DEF = 75^\circ.$ Therefore, our answer is $\angle DAC + \angle ADE = 80^\circ + 75^\circ = \boxed{155^\circ}.$