Problems that require the search for a proof of a particular theorem are common in a subject such as geometry.
One of them is the proof of the equality of a segment and a bisector.
You will need
- - notebook-
- - a pencil-
- - ruler.
It is impossible to prove the theorem without knowing itConstituents and their properties. It is important to pay attention to the fact that the bisector of the angle, in accordance with the generally accepted concept, is a ray emerging from the top of the corner and dividing it by two more equal angles. In this case, the angle bisector is considered to be a special geometric location of points inside the corner, which are equidistant from its sides. According to the proposed theorem, the angle bisector also represents a segment that emerges from the angle and intersects with the opposite side of the triangle. This assertion must be proved.
Familiarize yourself with the notion of a segment. In geometry, this is part of a straight line, bounded by two or more points. Given that the point in geometry is an abstract object without any characteristics, we can say that the segment is the distance between two points, for example, A and B. Points bounding a segment are called its ends, and the distance between them is its length.
Proceed to the proof of the theorem. Formulate its detailed condition. For this, we can consider a triangle ABC with a bisector BK issuing from the angle B. Prove that BK is a segment. Through the vertex C, draw a straight line CM that runs parallel to the bisectrix VC until it intersects with the side AB at the point M (for this, the side of the triangle needs to be continued). Since VC is the bisector of the angle ABC, it means that the angles AVK and KBC are equal to each other. Also the corners of the AVK and BMC will be equal because they are the corresponding angles of two parallel straight lines. The next fact is the equality of the angles of the CVC and the BCM: these are the angles lying in the opposite direction with parallel straight lines. Thus, the angle of the BCM is equal to the angle of the IUD, and the IUD triangle is isosceles, so BC = BM. Guided by the theorem on parallel lines that intersect the sides of the angle, you get the equality: AK / KC = AB / BM = AB / BC. Thus, the bisector of the inner corner divides the opposite side of the triangle by proportional parts to its contiguous sides and is an interval, which was to be proved.