Question 1038355
{{{drawing(400,4400/21,-1,20,-1,10,
locate(0,0,A),locate(15.3,0,C), locate(5,9.3,B),triangle(0,0,1,0,23,0),
triangle(0,0,10cos(60*pi/180),10sin(60*pi/180),15.32088886,0),
locate(1,1,"60°"), locate(5,7.3,"80°"), red(arc(0,0,6,-6,0,60),
arc(10cos(60*pi/180),10sin(60*pi/180),5,-5,240,320),
arc(15.3,0,4,-4,0,140)),locate(19.6,0,D), locate(15,1.2,"???°")
 )}}}<b><pre>

The measure of an exterior angle of a triangle is equal to 
the sum of the measures of the two remote interior angles.
&#8736;A and &#8736;B are interior angles of &#8710;ABC and &#8736;BCD is an
exterior angle of &#8710;ABC.   

Therefore:  m&#8736;BCD = m&#8736;A + m&#8736;B = 60° + 80° = 140° 

Edwin</pre></b>