Prove that the sum of the three angles of a triangle is 180 degrees.
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TRIANGLE
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.
Definition
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'.
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.
Definition
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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DevasmitaDutta:
hope it helps.....:D
Answered by
1
Sum of the angles of a polygon = (2n-4)90°
where n is the number of sides
Thus ,the sum of angles of a triangle. = (2×3-4)90°
=(6-4)90°
= 2×90° = 180°
where n is the number of sides
Thus ,the sum of angles of a triangle. = (2×3-4)90°
=(6-4)90°
= 2×90° = 180°
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