Question 832709
<pre>
You did not tell us which angles are &#8736;1 and &#8736;2.  So I'll guess
they are alternate interior angles.  But whatever they are, the
gist of the problem is similar.

{{{drawing(400,4400/19,-14,5,-4,7,locate(1.4,3.4,1),locate(5,0,b),
locate(1.3,.8,2), locate(.6,.8,3),locate(3,6,t),locate(3.5,4,a),
line(0,-3,3,6), line(0,3,4,4),line(-2,0,5,0) )}}}

Assume for contradiction that a and b are not parallel.
Then if we extend them far enough, they will meet at some 
point P.

{{{drawing(400,4400/19,-14,5,-4,7,locate(1.4,3.4,1),locate(5,0,b),
locate(1.3,.8,2), locate(.6,.8,3),locate(3,6,t),locate(3.5,4,a),
line(0,-3,3,6), line(0,3,4,4),line(-2,0,5,0),
green(line(-12,0,0,3),line(-12,0,-2,0), locate(-12,0,P))

 )}}}

m&#8736;1 = m&#8736;2  because measure of congruent angles are equal.

m&#8736;2 + m&#8736;3 = 180° because they form a linear pair

m&#8736;1 + m&#8736;3 = 180°  substituting equals for equals

m&#8736;P + (m&#8736;1 + m&#8736;3) = 180°  the measures of the three angles of a triangle
                         have sum 180°

m&#8736;P + 180° = 180°  substituting 180° for (m&#8736;1 + m&#8736;3), equals for equals.

m&#8736;P = 0°  subtracting equals from equals (180°=180°)

A triangle cannot have an angle with 0 measure.

Therefore we have reached a contradiction to the assumption that a and b
are not parallel.

Therefore the assumption is incorrect and therefore a&#8741;b

Edwin</pre>