Question 713497
<pre>
Since &#8736;A = 20° and &#8736;B = 80°
&#8736;ACB = 180°- 20° - 80° = 80°

Therefore &#5123;ABC is isosceles with AB = AC. 

{{{drawing(500,300,-1,4,-1,2,
line(2.75,.484,2.83,.5),
triangle(0,0,2.879385242,0,2.705737064,.984807753),
triangle(0,0,1,0,2.705737064,.984807753),
locate(0,0,A),locate(2.88,0,B), locate(2.7,1.1,C),
locate(1,0,D),locate(.43,.18,"20°"), locate(2.6,.2,"80°"),
line(.5,-.05,.5,.05) )}}}

Locate point P so that &#5123;APD &#8773; &#5123;ABC,
This is possible because AD = BC.  Also draw in PC:
{{{drawing(500,500,-1,4,-1,4,
line(2.75,.484,2.83,.5), line(.5,-.05,.5,.05),
green(line(2.879385242cos(80*pi/180),2.879385242sin(80*pi/180),2.705737064,.984807753)),
red(triangle(2.879385242cos(80*pi/180),2.879385242sin(80*pi/180),1,0)),
triangle(0,0,2.879385242,0,2.705737064,.984807753),
triangle(0,0,1,0,2.705737064,.984807753),
locate(0,0,A),locate(2.88,0,B), locate(2.7,1.1,C),
locate(1,0,D),locate(.40,.16,"20°"), locate(2.6,.2,"80°"), locate(.4,2,"20°"),
locate(.1,.3,"60°"),locate(.75,.16,"80°"),locate(.65,2.5,"40°"),
locate(1,.25,"70°"), locate(.5,3,P)

 )}}}

I will now explain all those angle measures 
I've written in:

AB = AC = PA = PD by construction 

&#8736;PAC = 60° because &#8736;DAP = 80° and &#8736;CAD = 20°, 
and &#8736;PAC is their difference.

&#5123;PAC is isosceles because PA=AC and
since &#8736;PAC = 60°, &#5123;PAC is equilateral.

Therefore all the longest lines are
equal. AB=AC=PA=PD=PC

All angles of an equilateral triangle are 60°, 
so &#8736;APC is 60° and thus &#8736;DPC=40° since &#8736;APD=20° 
and &#8736;DPC is the difference between a 60° angle 
and a 20° angle.

&#5123;DPC is isosceles with a vertex angle of 40°, 
therefore its equal base angles &#8736;PDC and &#8736;PCD 
are 70° each.

Therefore &#8736;ADC = 80°+70° = 150°

Answer &#8736;ADC = 150°

Edwin</pre>