The properties of the corners at the base of an isosceles triangle are an important topic, in particular, it helps people who decide to be architects or engineers. The construction of the correct drawings is a necessary component of such professions. It is important that even drawing is based on knowledge of these properties, as they help to draw the correct proportions.
Corner properties at the base of an isosceles triangle
The first theorem is based on the assertion that the angles adjacent to the base of the triangle are the same in degree measure. The second theorem is based on the fact that in a triangle of this kind of bisector, which is perpendicular to the base, it can be considered the median and height.
Hence the third theorem. It says that the median, which is drawn to the base of a given triangle, can simultaneously be a height and a bisector. And, of course, the fourth theorem states that the height, which is drawn perpendicular to the base, is considered the median and bisector.
It is important to always remember the properties of the corners at the base and the definition of an isosceles triangle, which states that such a figure has sides equal in length to each other.
Proof of
As an example of the proof of the theorem, we can consider the isosceles triangle ABC, which has a lower side BC. It is necessary to prove that the angle B is equal to the angle C. We can construct a bisector with the notation AD. It causes a series of sequences, as it divides one triangle into two identical ones. They are the same, because this is the first sign of the equality of triangles (they have a common side). Thus, the angle B will be equal to the angle C. As required.
One more theorem is deduced from this proved property of angles at the base of an isosceles triangle. It concerns the third sign of the equality of triangles.
If the three sides of one triangle are respectively equal to the three sides of another triangle, then such triangles are equal.
Before embarking on an example, it is important to understand the following. There is a concept of mid-perpendiculars that intersect at a specific point if drawn to the sides of the triangle.
Examples
It is necessary to prove with the help of existing knowledge that each point of the middle perpendicular is equally distant from the ends of the segment. Draw a perpendicular e, which will reach the segment AB. Point O will be the corresponding midpoint of AB.
We can consider the point L, which will be on the line e. Then make segments AL and BL. The resulting triangles are equal in total, because their angles at the vertex O are straight, OL will be a common leg, and the leg of OA will be OB. From the equality of the triangles it is clear that AL = BL. Q.E.D.