PROVE OF THE SUM OF THE INTERIOR ANGLES OF A TRIANGLE

PROVE OF THE SUM OF THE INTERIOR ANGLES OF A TRIANGLE

Taking as a reference this clothes hanger, whose inner edge is shaped like a triangle, we will call its internal angles as α, β y θ.

* Remember that α, β, y θ are part of the Greek Alphabet, composed by twenty-four letters, used to write the greek language.

We draw a parallel line through B, so that we can apply the property of the alternate interior angles.

* In the alternate interior angles it is established that when two parallel lines are cut by a transversal,  the alternate interior angles that result are equals.

We obtain from the clothes hanger's inner contour a triangle that we will call ABC according to its vertex points.

We draw a parallel line L₁ to the side AC of the triangle ABC in the vertex point B, so that we can apply the theorem of the alternate interior angles between parallel lines.

*If two parallel lines (L₁ y AC) are cut by a transversal (BC o AB), then the pairs of alternate interior angles are equals.

We see in the vertex point B that a straight angle or flat angle has been formed with the angles α, β y θ, from which the sum of all the angles is 180º.

In that sense, it is proven that α + β + θ = 180º, angles that form all the interior angles of the triangle.

* The side AC of the triangle acts as another parallel line to L1.

* The parallel line L1, is a imaginary line that has been drawn so that it facilitates its visualization with segment line or side AC of the triangle, so that we can profit the equidistant distance (the same distance) between parallel lines.