SOLUTION: Please help me solve this question: Four angles of a pentagon are equal and the fifth is 60°. Find the angles and show that two sides of the pentagon are parallel. Thank you

The pentagon ABCDE above is given to be drawn such that

∠A = 60°, and ∠B = ∠C = ∠D = ∠E 

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

   (n-2)∙180°

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

   (5-2)∙180° =
     (3)∙180° =
         540°

So  ∠A + ∠B + ∠C + ∠D + ∠E = 540°

Suppose ∠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,

∠A = 60°,  ∠B = ∠C = ∠D = ∠E = 120° 

Now we recall a well-known theorem of geometry:

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°).

∠A and ∠B are supplementary because ∠A + ∠B = 60° + 120° = 180°

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

Therefore by the theorem, AE ∥ BC.

Similarly,

∠A and ∠E are supplementary because ∠A + ∠E = 60° + 120° = 180°

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

Therefore by the theorem, AB ∥ EC.

Edwin