The sum of the corners of the triangle

Theorem.

The sum of the angles of a triangle is 180º.



Evidence. Let triangle ABC be given. Let us draw a straight line through vertex B parallel to AC. We mark on the obtained straight line point D so that it lies in the other half-plane with respect to the straight line BC.
∠ CAB and ∠ ABD - internal one-sided angles for parallel straight lines AC and BD with secant AB, then:
∠ CAB + ∠ ABD = 180º ⇒ ∠ ABD = 180º - CAB
∠ ABD = ∠ ABC + ∠ CBD.
Since ∠ CBD = ∠ ACB as internal crosswise lying, formed by the intersection of parallel straight lines BD and AC c section BC, then
∠ ABD = ∠ ABC + ∠ ACB
Equate ∠ ABD:
∠ ABC + ∠ ACB = 180º - ∠ CAB
And ∠ ABC + ∠ ACB + ∠ CAB = 180º
The theorem is proved.

From the theorem it follows:
Any triangle has at least two acute angles.
Evidence.
Suppose that a triangle has one acute angle or no angle at all. Then, at least, this triangle has two obtuse angles. And the degree measure of the dull angle is more than 90º. So the sum of two obtuse angles will already be more than 180º. And this is impossible, since the sum of all angles of the triangle is 180º. Q.E.D.