Question 969188
<pre>
{{{drawing(400,2000/9,-1,8,-1,4,

line(0,0,2,3),
line(7,3,2,3),
line(5,0,0,0),
line(5,0,7,3),
line(2,0,2,3),
line(5.461538462,.6923976923,2,3),

locate(-.1,0,B),
locate(1.9,0,E),
locate(5,0,C),
locate(5.55,.8,F),
locate(7,3.4,D),
locate(2,3.4,A)





)}}}

Given: parallelogram ABCD, AE &#8869; BC, AF &#8869; CD. 
To prove: &#916;ABE &#8765; &#916;ADF.

1. m&#8736;AEB = 90°      Given AE &#8869; BC.
2. m&#8736;AFD = 90°      Given AF &#8869; CD. 
3.  &#8736;AEB &#8773; &#8736;AFD    Both have equal measures. 1,2
4.    &#8736;B &#8773; &#8736;D      Opposite interior &#8736;s of parallelogram ABCD are &#8773;.
5.  &#916;ABE &#8765; &#916;ADF    If 2 &#8736;s of one &#916; are &#8773; 2 &#8736;s of another &#916; the &#916;s are &#8765;. 3,4

Edwin</pre>