Tasks that provide proof search of a theorem, common in such subjects as geometry.
One of them is the proof of the equality of the segment and bisector.
You will need
- - notebook-
- - karandash-
- - Line.
Prove the theorem is not possible without knowledge of itscomponents and their properties. It is important to note that the bisector of the angle, according to the conventional concept, is a ray emanating from the vertex of the angle and divides it into two equal angles. This bisector is considered a special locus location inside the angle, which are equidistant from its sides. According to the nominated theorem, the angle bisector of the segment also is coming out of a corner and intersects the opposite side of the triangle. This statement and should prove.
Check out the segment concept. The geometry of this part of the line, bounded by two or more points. Given that a point in the object is abstract geometry without any characteristics, it can be said that the segment - the distance between two points, e.g., A and B. The points that limit the section called its ends, and the distance between them - of its length.
Proceeding to the proof of the theorem. Formulate a detailed condition. To do this, you can consider a triangle ABC the bisector BK, coming out of the corner of V. Prove that BK is a segment. A top with draw a straight CMS, which will be held parallel to the bisector of the VC to the intersection with the side AB at point M (for this side of the triangle should be continued). Since VC is the bisector of the angle ABC, then the angles AVK and FAC are equal. It will also be equal to the angles of the AVK and Navy because it corresponding angles of two parallel lines. The next fact is the equality of the angles KBC and SCM: the angles, lying crosswise with parallel lines. Thus, SCM angle equals the angle of the Navy, and the Navy triangle is isosceles, so BC = BM. Guided by the theorem of parallel lines that cross hand corner, you'll get equality: AK / COP = AB / BM = AB / BC. Thus, the internal bisector of the angle divides the opposite side of the triangle to its proportional part of the adjacent side segment and is, as required.