Prove that the sum of the angles of Triangle is 180°.

Too messy to type here.

That proof should be in most textbooks, if not,
Google it.

https://search.yahoo.com/search?fr=spigot-nt-gcmac&ei=utf-8&ilc=12&type=692858&p=proof%20:%20sum%20of%20angles%20in%20a%20triangle%20equals%20180

To prove that the sum of the angles of a triangle is 180°, we can use the properties of parallel lines and angles.

1. Draw a triangle ABC.
2. Draw a line segment DE parallel to the side BC, passing through the vertex A. In other words, line DE should not intersect any of the sides of the triangle ABC.
3. Now, we have two transversal lines (AB and DE) crossing three parallel lines (BC, DE, and AC).
4. By the property of alternate interior angles, the angles ∠ABC and ∠CDE are congruent (marked as ∠1).
5. By the property of corresponding angles, the angles ∠BAC and ∠CED are congruent (marked as ∠2).
6. By the property of alternate interior angles, the angles ∠ACB and ∠EDC are congruent (marked as ∠3).

Now, we have three pairs of congruent angles: ∠1, ∠2, and ∠3. Since the sum of all angles at a point is 360°, we can represent it as:

∠1 + ∠2 + ∠3 + ∠EDC = 360°

Substituting the corresponding angles:

∠ABC + ∠BAC + ∠ACB + ∠EDC = 360°

Since ∠EDC is congruent to ∠ACB:

∠ABC + ∠BAC + ∠ACB + ∠ACB = 360°

Combining like terms:

∠ABC + ∠BAC + 2∠ACB = 360°

Now, observe that ∠ABC + ∠BAC + ∠ACB is the sum of the angles in a triangle. Let's call it Σ (Sigma).

Σ + ∠ACB = 360°

Rearranging the equation:

Σ = 360° - ∠ACB

Finally, since ∠ACB is an angle of the triangle, Σ represents the sum of all angles in a triangle. The equation can be written as:

Σ = 180°

Therefore, the sum of the angles in a triangle is proven to be 180°.