Question 1021212
<pre>
{{{drawing(400,400,-7,7,-4,10,

line(-3.845299462,-2,-5,0),
line(-3.845299462,-2,-.7823646297,-2),
line(-.7823646297,-2,2.35,3.007674664),
line(2.35,3.007674664,0,5sqrt(3)),
line(0,5sqrt(3),-5,0),
locate(-3.85,-2,C),locate(-.78,-2,D),locate(2.45,3.42,E),
locate(-.25,9.25,A),locate(-5.38,.39,B)


 )}}}

The pentagon ABCDE above is given to be drawn such that

&#8736;A = 60°, and &#8736;B = &#8736;C = &#8736;D = &#8736;E 

The sum of the interior angles of a polygon of n sides
is given by the formula:

   (n-2)&#8729;180°

The pentagon above has 5 sides, so n=5.  Substituting n=5,

   (5-2)&#8729;180° =
     (3)&#8729;180° =
         540°

So  &#8736;A + &#8736;B + &#8736;C + &#8736;D + &#8736;E = 540°

Suppose &#8736;A = 60° and all the other angles = x°
then:

    60° + x° + x° + x° + x° = 540°
                  60° + 4x° = 540°

Subtract 60° from both sides:

                        4x° = 480°

Divide both sides by 4

                         x° = 120°

So in the figure above,

&#8736;A = 60°,  &#8736;B = &#8736;C = &#8736;D = &#8736;E = 120° 

Now we recall a well-known theorem of geometry:
<b><i>
When a transversal intersects two lines, the two 
lines are parallel if and only if interior angles 
on the same side of the transversal are 
supplementary (sum to 180°).</b></i>

&#8736;A and &#8736;B are supplementary because &#8736;A + &#8736;B = 60° + 120° = 180°

Lines AE and BC are two lines cut by transversal AB.

Therefore by the theorem, AE &#8741; BC.

Similarly,

&#8736;A and &#8736;E are supplementary because &#8736;A + &#8736;E = 60° + 120° = 180°

Lines AB and EC are two lines cut by transversal AE.

Therefore by the theorem, AB &#8741; EC.

Edwin</pre>