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Given the figure as described, draw ````a line parallel to `bar(BC)` through A intersecting `bar(CD)` at E.
Now `bar(AB)||bar(CD)` and `bar(AE)||bar(BC)` by construction, so ABCE is a parallelogram. In a parallelogram, opposite sides are congruent so `bar(BC) cong bar(AE)` ; then `bar(AE) cong bar (AD)` by the transitive property of congruence. So `Delta AED` is isosceles, and `/_D cong /_ AED` . But `/_AED cong /_C` since they are corresponding angles of the parallel lines. Therefore `/_D cong /_C` .
Then `/_A cong /_B` since `/_A` is supplementary to `/_D` and `/_B` is supplementary to `/_C` and supplements of congruent angles are congruent.
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