Question

prove that sum of all sides of a triangle are 180°

Answer 1

Hey mate★★

Here is your answer★★

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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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★★★♥Thanks for giving me opportunity to help★★★♥

Answer 2

Mistake: not all sides.... It is all angles......
☺️
<a+<b+ <c = 180°
A triangle has 3 sides. Thus, three angles also.
So, if we take it an equilateral triangle. Then, every angles are of same value.
<a= <b= <c
Which is = 180÷3 = 60°
6 ×3 = 180°
Thus, proved........

Or,
According to the Theorem.... Sum of all angles of a triangle = 180°

HOPE it helps you