proof that the sum of the triangle is 180
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Proof that the sum of the angles in a triangle is 180 degrees
Theorem
If ABC is a triangle then <)ABC + <)BCA + <)CAB = 180 degrees.
Proof
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.
Lemma
If ABCD is a quadrilateral and <)CAB = <)DCA then AB and DC are parallel.
Proof
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.
Theorem
If ABC is a triangle then <)ABC + <)BCA + <)CAB = 180 degrees.
Proof
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.
Lemma
If ABCD is a quadrilateral and <)CAB = <)DCA then AB and DC are parallel.
Proof
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.
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Answered by
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Let the all angles be x
x + x + x = 180 °
3x = 180°
x = 180° / 3
x = 60 °
x + x + x = 180 °
3x = 180°
x = 180° / 3
x = 60 °
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