Question 1166662
Four angles of a pentagon are equal and the fifth is 60°. Find the equal
angles and show that two sides of the pentagon Are parallel.
<pre>
The sum of the n interior angles of an n-sided polygon is given by

(n-2)180°

A pentagon has 5 sides, so n = 5 and the sum of the interior angles is

(5-2)180° = (3)180° = 540°

The fifth angle is 60°, so the sum of the other 4 is 540°-60°=480°.
They are all equal, so each one is 480°/4 = 120° each

{{{drawing(400,1400/3,-1.5,1.5,-1.3,2.2,

line(-.5,-sqrt(3)/2,.5,-sqrt(3)/2), line(.5,-sqrt(3)/2,1,0),
line(1,0,0,sqrt(3)),line(-.5,-sqrt(3)/2,-1,0),
line(-1,0,0,sqrt(3)),

locate(-.04,1.88,A), locate(-1.08,.05,B), locate(-.5,-.87,C),
locate(1.03,.05,E), locate(.5,-.87,D),

locate(-.5,-.68,120^o), locate(.31,-.68,120^o),
locate(-.96,.13,120^o), locate(.7,.13,120^o), locate(-.08,1.6,60^o) )}}}

∠A, ∠B are supplementary because their measures are 60° and 120°
which have sum 180°

AE ∥ BC because consecutive interior angles A and B are supplementary, when
AE and BC are cut by transversal AB.

AB ∥ ED similarly.

Edwin</pre>