Question 727836
{{{drawing(300,300,-12,12,-12,12,
line(8.66,-10,8.66,15),line(-8.66,-10,-8.66,0),
line(-8.66,0,0,5),line(0,5,8.66,0),
line(-8.66,-10,0,-15),line(0,-15,8.66,-10),
blue(line(-8.66,0,8.66,0)),
green(line(0,5,15,12.99)),
locate(-9.5,0.5,A),locate(9,0.5,C),
locate(-0.4,6.2,B),locate(9,10,G),
locate(-9.5,-9.5,F),locate(9,-9.5,D)
)}}} Hopefully exterior angle 1 is GCB. If side CD was extended in the other direction, I had to do more drawing, but the idea is the same.
The exterior angles of a polygon are the changes in direction as you move around along the perimeter. Going counterclockwise, you go from D to C, and change direction so that instead of continuing straight towards G, you head to B. The angle GCB is an exterior angle.
If you go around clockwise, you change direction at B, and instead of heading towards G, to go towards C. Angle GBC is an exterior angle too.
You can draw 2 exterior angles at each vertex.
Because a regular polygon is symmetrical, in an regular polygon all the exterior angles are congruent. That is why it does not matter which of the exterior angles was labeled as angle 1. They all have the same measure as GCB and GBC.
Angle GBC in supplementary to angle ABC, so
{{{m<ABC+m<GBC=180^o}}}
In triangle ABC, with BAC=2 and BCA=3,
{{{180^o=m<A}}}{{{BC + m<BAC + m<BCA}}}
The 2 equations together result in
{{{m<ABC + m<G}}}{{{BC=m<A}}}{{{BC + m<BAC + m<BCA }}} --> {{{m<GBC = m<B}}}{{{AC + m<BCA}}}
And since all exterior angles have the same measure,
{{{m<1 = m<G}}}{{{CB=m<GBC=m<B}}}{{{AC+m<B}}}{{{CA=m<2+m<3}}}