Well, the point on the ellipse (x,y) is:

x = a cos t
y = b sin t

Where a is half the ellipse's width, b is half the ellipse's height, and t varies from 0 to TWO_PI.

So plug in a, b, and t to those equations to find the point for the other end of the line you want to draw.

That is:

void arcBound( float x, float y, float w, float h, float s, float e ){
  arc(x, y, w, h, s, e );
  line( x, y, x + w/2.0 * cos(s), y + h/2.0 * sin(s) );
  line( x, y, x + w/2.0 * cos(e), y + h/2.0 * sin(e) );
}

That's typical of how duplicate threads can make two people to do the same job.
I deleted the duplicate thread, but avoid to do this in the future.
Just not to loose my time, I paste my answer here, even though tfguy44 already gave the answer.

Reusing the PieChart example of Processing:

1. size(400, 400);
2. background(200);
3. smooth();
4. //noStroke();
5. stroke(#886600);
6. int centerX = width / 2;
7. int centerY = height / 2;
8.  
9. float diameter = min(width, height) * 0.75;
10. int[] angs = { 30, 10, 45, 35, 60, 38, 75, 67 };
11. float lastAng = 0;
12.  
13. for (int i = 0; i < angs.length; i++)
14. {
15.   fill(angs[i] * 3.0);
16.   float newAng = radians(angs[i]);
17.   arc(centerX, centerX, diameter, diameter, lastAng, lastAng + newAng);
18.   line(centerX, centerY, centerX + diameter/2 * cos(lastAng), centerY + diameter/2 * sin(lastAng));
19.   lastAng += newAng; 
20. }

Thanks everyone - this helped out a lot!