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Proof that the sum of the angles in a triangle is 180 degrees

If ABC is a triangle then <)ABC + <)BCA + <)CAB = 180 degrees.

Draw line a through points A and B. Draw line b through point C and parallel to line a.

Since lines a and b are parallel, <)BAC = <)B'CA and <)ABC = <)BCA'.
It is obvious that <)B'CA + <)ACB + <)BCA' = 180 degrees.
Thus <)ABC + <)BCA + <)CAB = 180 degrees.

If ABCD is a quadrilateral and <)CAB = <)DCA then AB and DC are parallel.

Assume to the contrary that AB and DC are not parallel.
Draw a line trough A and B and draw a line trough D and C.
These lines are not parallel so they cross at one point. Call this point E.

Notice that <)AEC is greater than 0.
Since <)CAB = <)DCA, <)CAE + <)ACE = 180 degrees.
Hence <)AEC + <)CAE + <)ACE is greater than 180 degrees.
Contradiction. This completes the proof.

Two Triangles ABC and A'B'C' are congruent if and only if
|AB| = |A'B'|, |AC| = |A'C'|, |BC| = |B'C'| and,
<)ABC = <)A'B'C', <)BCA = <)B'C'A', <)CAB = <)C'A'B'.

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