Astraight line that is parallel to one of the sides of a given triangle intersects its other two sides (or their extensions) and forms a triangle together with them. prove that this triangle is similar to the original triangle.

We know that this triangle is similar because it will have all of the same angles as the first triangle. This is because two parallel lines cut by a transversal will create the same angles. In addition, they share the final angle because we are not changing that angle. Therefore, all 3 are the same, which makes them similar triangles.

In a general context (because there's no specific data in the problem), we could demonstrate the similarity using one of the postulates. Specifically, we could use the Angle-Angle postulate.

We have a triangle ABC, and inside of it there's a line that crosses to sides, and its parallel to the third side. This set-up gives us parallels that are being intersected by transversal, this means that all angles are congruent between the triangle ABC and the new triangle DBE.

Basically, they share the same upside angle, and the other two at their base are also congruent because they are corresponding angles.

Therefore, by AA postulate, both triangles are similar.

2

step-by-step explanation:

my answer is$ 12

step-by-step explanation:

because 15 -2=12