Triangle Problem from /b/

http://i29.tinypic.com/i40e8n.jpg

Method for obtaining a unique solution:

Assign an arbitrary length to the equal sides.  L.
Assume the bottom left corner is at the origin and derive the equations of the lines to find the endpoints of DE.
Turn DE and AE into vectors and take the dot product.

Line AD:
y = tan(80)*x

Line BD:
(2LSin10, 0)
m = -tan(60)
y = -tan60 (x - 2LSin10)

Point D:
tan(80)x = -tan60 (x - 2LSin10)
(tan80 + tan60)x = 2Lsin(10)tan(60)
x = 2Lsin(10)Tan(60) / (tan80 + tan60)
y = 2Lsin(10)Tan(60)Tan(80) / (tan80 + tan60)

Line AE:
y = tan(70)*x

Line BE:
(2LSin10, 0)
m = -tan(80)
y = -tan(80) (x - 2LSin10)

Point E:
tan(70)*x = -tan(80) (x - 2LSin10)
(tan70 + tan80)*x = 2LSin(10)Tan(80)
x = 2LSin(10)Tan(80) / (tan70 + tan80)
y = 2LSin(10)Tan(80)Tan(70) / (tan70 + tan80)

Vector DE:
x: 2Lsin(10)Tan(60) / (tan80 + tan60) - 2LSin(10)Tan(80) / (tan70 + tan80)
y: 2Lsin(10)Tan(60)Tan(80) / (tan80 + tan60) - 2LSin(10)Tan(80)Tan(70) / (tan70 + tan80)

Each term has 2Lsin(10) which can be factored out, Leaving:
<.43969,.52400>

Vector AE:
x: 2LSin(10)Tan(80) / (tan70 + tan80) - 0
y: 2LSin(10)Tan(80)Tan(70) / (tan70 + tan80) - 0

Again factoring out 2Lsin(10) leaves us with:
<.63736,1.8508>

DE (dot) AE = |DE| * |AE| * Cos(x)
1.2500 = 1.3389 * Cos(x)
Cos(x) = .93354
x = 21

>>1
Entire point was to do it with basic geometry. It's trivial with trig.