Given tri ABC with <B=>C=80 degrees and >A=20 degrees  Isosceles!  Draw angle ABD=20 degrees so that D is a point on side AC. Draw angle ACE=10 degrees so that E is a point on side AB. Let BD and CE intersect at T  Draw line segment DE. Required to prove that Triangle AED is similar to Triangle BTE. Let angle ADE=x.   I can do this proof by imposing a measurement of Sin 20 on line segment DE and using the sine rule to establish simple expressions for AE,AD,BD BT and BE.  Then ratios of corresponding sides leads to angle ADE = 30 degrees hence AA similarity. (Sin20/Sin20=1)

Is this the easiest method. Can I do a geometric extension to the diagram to establish angle ADE using parallel lines or circles and tangent or angles in circles....I have tried them all but to no avail.  Help please.