Since ∠A = 20° and ∠B = 80°
∠ACB = 180°- 20° - 80° = 80°
Therefore ᐃABC is isosceles with AB = AC.
Locate point P so that ᐃAPD ≅ ᐃABC,
This is possible because AD = BC. Also draw in PC:
I will now explain all those angle measures
I've written in:
AB = AC = PA = PD by construction
∠PAC = 60° because ∠DAP = 80° and ∠CAD = 20°,
and ∠PAC is their difference.
ᐃPAC is isosceles because PA=AC and
since ∠PAC = 60°, ᐃPAC is equilateral.
Therefore all the longest lines are
equal. AB=AC=PA=PD=PC
All angles of an equilateral triangle are 60°,
so ∠APC is 60° and thus ∠DPC=40° since ∠APD=20°
and ∠DPC is the difference between a 60° angle
and a 20° angle.
ᐃDPC is isosceles with a vertex angle of 40°,
therefore its equal base angles ∠PDC and ∠PCD
are 70° each.
Therefore ∠ADC = 80°+70° = 150°
Answer ∠ADC = 150°
Edwin