SOLUTION: Triangle ABC is such that ∠A=20∘ and ∠B=80∘. The point D on line segment AB satisfies AD=BC. What is the measure (in degrees) of ∠ADC?

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Question 713497: Triangle ABC is such that ∠A=20∘ and ∠B=80∘. The point D on line segment AB satisfies AD=BC. What is the measure (in degrees) of ∠ADC?
Answer by Edwin McCravy(20055)   (Show Source): You can put this solution on YOUR website!
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
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