Hi Rachel,
Suppose the triangle has vertices ABC, then extend one side (I extended BC) to a point D. Next draw a line segment through C parallel to the opposite side (here AB).
Since AB and EC are parallel, and AC is a transversal, angle CAB is congruent to angle ACE.
Since AB and EC are parallel, and BC is a transversal, angle ABC is congruent to angle ECD.
Thus
angle BCA + angle CAB + angle ABC = angle BCA + angle ACE + angle ECD
But
angle BCA + angle ACE + angle ECD = angle BCD = 180^{o}
Thus
angle BCA + angle CAB + angle ABC = 180^{o}
Penny
