The tasks providing search of the proof of any given theorem are widespread in such subject as geometry. One of them is the proof of equality of a piece and bisector.
It is required to you
- - notebook;
- - pencil;
- - ruler.
Instruction
1. It is impossible to prove the theorem without knowledge of its components and their properties. It is important to pay attention that the bisector, according to the standard concept, represents the beam leaving vertex of angle and dividing it into two more equal corners. At the same time the bisector is considered a special locus of arrangement of points in a corner which are equidistanted from its parties. According to the put-forward theorem, the bisector also represents the piece leaving a corner and which is crossed with the opposite side of a triangle. This statement also should be proved.
2. Get acquainted with a concept of a piece. In geometry it is the part of a straight line limited to two or more points. Considering that the point in geometry is an abstract object without any characteristics, one may say, that a piece – distance between two points, for example, of A and B. The points limiting a piece are called its ends, and distance between them - its length.
3. Start the proof of the theorem. Formulate its detailed condition. For this purpose it is possible to consider AVS triangle with a bisector of BK leaving V. Dokazhite's corner that BK is a piece. Through top With draw the CM straight line which will pass parallel to VK bisector before crossing with the party of AV in the M point (for this purpose the party of a triangle needs to be continued). As VK is AVS bisector, so corners of ABK and KBC are equal among themselves. Corners of AVK and Naval Forces because it is corresponding corners of two parallel straight lines will be also equal. The following fact consists in equality of corners of KBC and BCM: these are the corners lying crosswise at parallel straight lines. Thus, the corner of VSM is equal to a corner of Naval Forces, and the triangle of Naval Forces is isosceles therefore BC=BM. Being guided by the theorem of parallel straight lines which cross sides of angle, you receive equality: AK/KC=AB/BM=AB/BC. Thus, the bisector of an internal corner divides the opposite side of a triangle into parts proportional to its adjacent parties and is a piece, as was to be shown.