Question 1089450
AB=BC=(1/2)AC because B is the midpoint of AC,
So, if AC=12, then AB=BC=6.
BD=DC=(1/2)BC because D is the midpoint of BC,
and because BC=6, BD=DC=3.
So, the line, with points A, B, C, and D,
and the distances between them, looks like the drawing below (or its mirror image).
{{{drawing(800,100,-2,14,-1,1,
line(-2,0.0,0),red(line(0,0,6,0)),
green(line(6,0,9,0)),blue(line(9,0,12,0)),
line(12,0,14,0),circle(0,0,0.05),
circle(6,0,0.05),circle(9,0,0.05),
circle(12,0,0.05),locate(-0.05,-0.08,A),
locate(5.95,-0.08,B),locate(8.95,-0.08,D),
locate(11.95,-0.08,C),locate(2.95,0,red(6)),
locate(7.45,0,green(3)),locate(10.45,0,blue(3))
)}}} .
 
We do not know exactly where point E is located on that line,
but we know that DE=2.
 
It could be like the drawing below (or its mirror image).
{{{drawing(800,100,-2,14,-1,1,
line(-2,0.0,0),red(line(0,0,6,0)),
green(line(6,0,9,0)),blue(line(9,0,12,0)),
line(12,0,14,0),circle(0,0,0.05),
circle(6,0,0.05),circle(9,0,0.05),
circle(12,0,0.05),locate(-0.05,-0.08,A),
locate(5.95,-0.08,B),locate(8.95,-0.08,D),
locate(11.95,-0.08,C),locate(2.95,0,red(6)),
locate(7.45,0,green(3)),locate(10.45,0,blue(3)),
circle(7,0,0.05),locate(6.9,0.4,E),
arrow(9,0.1,7,0.1),arrow(7,0.1,9,0.1),
locate(7.95,0.5,2)
)}}} .
In that case,
AE=AB+BD-DE=6+3-2=7 .
 
Otherwise, it would be like the drawing below (or its mirror image).
{{{drawing(800,100,-2,14,-1,1,
line(-2,0.0,0),red(line(0,0,6,0)),
green(line(6,0,9,0)),blue(line(9,0,12,0)),
line(12,0,14,0),circle(0,0,0.05),
circle(6,0,0.05),circle(9,0,0.05),
circle(12,0,0.05),locate(-0.05,-0.08,A),
locate(5.95,-0.08,B),locate(8.95,-0.08,D),
locate(11.95,-0.08,C),locate(2.95,0,red(6)),
locate(7.45,0,green(3)),locate(10.45,0,blue(3)),
circle(11,0,0.05),locate(10.9,0.4,E),
arrow(9,0.1,11,0.1),arrow(11,0.1,9,0.1),
locate(9.95,0.5,2)
)}}} .
In that case,
AE=AB+BD+DE=6+3+2=11 .
 
The sum of the possible lengths of segment AE is
{{{7+11=highlight(18)}}} .
 
If we did not know the distance AC, but we knew that DE<DC=BC, we would know that the points are in one of the two orders shown, and
AE=AB+BD-DE or AE'=AB+BD-DE ,
so AE+AE'=AB+BD-DE+AB+BD+DE=AB+BD+AB+BD=AD+AD=2AD .