Let's draw a triangle ABC:



Now through C, draw a line segment DE parallel to AC.


The two angles indicated by red arcs, ∠ACD and ∠BAC, are equal in measure 
because they are alternate interior angles when the transversal AC cuts
the two parallel lines AB and DE.

Also the two angles indicated by green arcs, BCE and ABC, are equal in
measure because they are alternate interior angles when the transversal
BC cuts the two parallel lines AB and DE.

The sum of the measures of the three angles at the top have sum 180° 
because they form a straight angle. That is

m∠ACD + m∠ACB + m∠BCE = 180°  because they form straight angle ∠DCE

and since m∠ACD = m∠BAC, and  m∠BCE = m∠ABC, we can substitute
equals for equals and get

m∠BAC + m∠ACB + m∠ABC = 180°.

That proves it.  But it's only an outline.  You should write it up in a 
two-column proof.

Edwin