Question 824378
<pre>
Let's sketch a few sides of the regular polygon and extend
two alternate sides until they meet at a right angle &#8736;C.

{{{drawing(300,300,-2,2,-2,2,
line(1.5,0,1.06066017,1.06066017),


line(1.06066017,1.06066017,0,1.5),


line(0,1.5,-1.06066017,1.06066017),

green(line(1.06066017,1.06066017,.75,1.810660172),
line(0,1.5,.75,1.810660172)),

locate(0,1.75,A),locate(.75,1.95,C),locate(1.1,1.2,B),
locate(-1.2,1.2,D),locate(1.55,.1,E),





line(1.06066017,-1.06066017,1.5,0)
)}}}


Suppose the regular polygon has n sides.

&#8736;DAB &#8773; &#8736;EBA because interior angles of 
a regular polygon are congruent.

&#8736;CAB &#8773; &#8736;CBA  because they are supplements of
congruent angles

&#8736;ACB is a right angle because it is given that
the extensions of DA and EB form a right angle.

&#916;ABC is an isosceles right triangle.

m&#8736;CAB = m&#8736;CBA = 45° because they are base angles
of an isosceles right triangle. 

&#8736;CAB, which has measure 45°, is an exterior angle of the polygon.

The sum of all n exterior angles of any polygon is 360°

All n exterior angles of a regular polygon are congruent.

Therefore 

n×45° = 360°
    n = {{{"360°"/"45°"}}}
    n = 8

So the regular polygon has 8 sides.  It is a regular octagon:

{{{drawing(300,300,-2,2,-2,2,
line(1.5,0,1.06066017,1.06066017),


line(1.06066017,1.06066017,0,1.5),


line(0,1.5,-1.06066017,1.06066017),

green(line(1.06066017,1.06066017,.75,1.810660172),
line(0,1.5,.75,1.810660172)),

locate(0,1.75,A),locate(.75,1.95,C),locate(1.1,1.2,B),
locate(-1.2,1.2,D),locate(1.55,.1,E),

line(-1.06066017,1.06066017,-1.5,0),
line(-1.5,0,-1.06066017,-1.06066017),
line(-1.06066017,-1.06066017,0,-1.5),
line(0,-1.5,1.06066017,-1.06066017),
line(1.06066017,-1.06066017,1.5,0)
)}}}

Edwin</pre>