Question

prof that triangle 3 side are equal 180​

Answer 1

Step-by-step explanation:

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Answer 2

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.