Question

Prove that sum of all angles of triangle is 180 degree.

Answer 1

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.

Answer 2

Statement : The sum of the angles of a triangle is 180°.

To prove : sum of the angles of ΔABC is 180°

Construction : Draw a line $lm$ parallel to BC.

Proof :

Since $lm$ || BC, we have ;

∠2 = ∠y ..... (i) . [Alternate interior angles]

Similarly,

∠1 = ∠z ....... (ii) . [Alternate interior angles]

Also, sum of angles at a point a on line $lm$ is 180°.

∴ ∠2 + ∠x + ∠1 = 180°

⇒ ∠y + ∠x + ∠z = 180°

⇒ ∠x + ∠y + ∠z = 180°

⇒ ∠A + ∠B + ∠C = 180°

Therefore,

Sum of all angles of a Δ is 180°