Given: triangle ABC. Prove: m<1 + m<2 + m<3 = 180

Question

 Translate ΔABC by vector AB  to form straight <ABB'  along the vector.  The isometric properties of translation preserve _________, thus m<1 = m<B'BC'.  Since <ABB'  is a straight angle, we know that m<2 + m<CBC' + m<B'BC' = __________°.  Translation also preserve ________,  there for ensuring that line AC parallel to line BC'. Since line AC parallel to line BC', m<3 = m<CBC' because alternate ________ angles are congruent. By making two ________ into the  straight angle relationship of  m<2 + m<CBC' + m<B'BC' = 180º we  arrive at the proof that m<1 + m<2 + m<3 = 180º.

Answer 1

I am assuming you have two parallel lines cut by a transversal.  Use the fact alternate interior angles and alternate exterior angles are the same, as well as corresponding angles are the same.  You will see that m<1, m<2, and m<3 are related to the three angles joined together that make a straight line.  Straight lines are always 180 degrees.