'Unplugged in Glacier' could be a movie title (Sleepless in ...). Hope you enjoyed your trip.
I am using H as the intersection of AE and BF. Here is what I have so far:
From the original problem you can determine that:
Triangle ABC is isosceles and the lower angles are 80* (use * as a degree symbol).
Angle ACB is therefore 20*.
Angle ADB is 50*.
Angle BEA is 40*.
Angle AEC is 140* and angle BDC is 130*.
Now draw a line parallel to AB from E to the opposite side and lable the intersection F. Triangle FCE is also isosceles with lower angles 80*.
Therefore angle FEA is 60*.
From the original problem, triangle AEC is also isosceles since angle CAE and angle ACE are both 20*. Therefore triangle BFC is also isosceles and congruent with triangle AEC.
Trapezoid AFEB is isosceles (if there is such a term for trapezoids) because sides AB and FE are parallel and angles FAB and EBA are both 80* and angles AFE and BEF are both 100*. Diagonal AE and the new diagonal BF form congruent interior triangles AHF and BHE (H is the point where the diagonals intersect.)
So triangle FHE is isosceles with sides FH and EH equal. And we know that angle FEA is 60*; therefore angle EFB is also 60* and triangle FHE is equilateral.