SOME NON-STANDART PROBLEMS OF ANALYTIC GEOMETRY

Аннотация

In this article, the solutions of some problems that belong to the analytical geometry part of geometry, are not included in the school geometry course and the higher geometry course, and are considered non-standard. That is, the solution to the problematic questions related to rectangles with a relatively high level of complexity was found analytically. The considered issues serve to develop the geometric worldview of students who want to master analytical geometry in depth. Keywords. non-standart problems, analytic geometry, rectangles, bissektor, parallelogram, The quadrilateral, fashion similarly, rhombus. Introduction. In this work, the problems related to the internal properties of a square, a square and a parallelogram were solved. The goal of finding a solution to the problems presented in the article in an analytical way is set, and for this, the task of using the necessary fundamental properties is assigned. The solved problems are completely non-standard and are relevant for the development of geometric imagination for students of mathematics in higher education. Difenation. A triangle is a simple closed curve or polygon which is created by three line-segments. In geometry, any three points, specifically non-collinear,form a unique triangle and separately, a unique plane [1:179]. The SAS Similarity Theorem. Given a correspondence be-tween two triangles. If two pairs of corresponding sides are proportional, and the included angles are congruent, then the correspondence is a similarity[1:189]. The ASA Similarity Theorem. If two angles and the included side of one triangle are congruent to two angles and the included side of a second triangle, then the two triangles are congruent[5:220]. Theorem-1. If a square is drawn externally on each side of a parallelogram, prove that: (a) The quadrilateral, determined by the centers of these squares, is itself a square; (b) The diagonals of the newly formed square are concurrent with the diagonals of the original parallelogram. ABCD Proof: (a) is parallelogram. Points 28 P , Q , R and S are the centres of the four squares, DAIJ, DCLK, and CBFE, respectively (fig. S-1). PA DR and AQ QD (each is one-half a diagonal). Also, ADC is supplementary to DAB, and IAH is supplementary to DAB (since AIDHABright angle). Therefore, ADCIAH . ILM-FAN VA INNOVATSIYA ILMIY-AMALIY KONFERENSIYASI in-academy.uz/index.php/si Since, 29 m  R D C  m  Q D A  m  H A P  m  Q A I  4 5 0 ,  R D Q   Q A P . Thus, R D Q  P A Q (s.a.s), and Q R  Q P . In a fashion similarly, it may be proved that Q P  P S and P S  R S . Therefore, P Q R S is a rhombus. RDQ PAQ DQRAQP; PQR DQA DQA Since, , therefore, (by addition). Since, - right angle, (S-1) and P Q R S a is square. PQRS (b) To prove that the diagonals of are concurrent with the diagonals of Parallelograms A B C D , we must prove that a diagonal of the square and a diagonal of the parallelogram bisect each other. In other words, we prove that the diagonals of the square and the diagonals of the parallelogram all share the same midpoint, (i.e., point O ).  B A C   A C D , and m  P A B  m  R C D  4 5 0 ; therefor,  P A C   R C A . Since,  A O P   C O R and A P  C R , A O P  C O R (s.a.a). Thus, A O  C O , and P O  R O . Since, D B passes through the midpoint of A C , and similarly Q S passes through the midpoint of P R , and since A C and P R share the same midpoint (i.e., O), we have shown that AC,PR,DB, and QS are concurrent (i.e., all pass through point O ). Theorem-3. In right △𝐴𝐵𝐶, with right angle at 𝐶,𝐵𝐷 = 𝐵𝐶,𝐴𝐸 = 𝐴𝐶, 𝐸̅̅̅𝐹̅ ⊥ 𝐵̅̅̅𝐶̅, and 𝐷̅̅̅𝐺̅ ⊥ 𝐴̅̅̅𝐶̅. Prove that 𝐷𝐸 = 𝐸𝐹 + 𝐷𝐺. (S-2) Proof. Draw 𝐶̅̅̅𝑃̅ ⊥ 𝐴̅̅̅𝐵̅, also draw 𝐶̅̅̅𝐸̅ and 𝐶̅̅̅𝐷̅ (Fig. S-4). 𝑚∠3 + 𝑚∠1+𝑚∠2 = 90° 𝑚∠3 + 𝑚∠1 = 𝑚∠4 (#5) By substitution, 𝑚∠4 +𝑚∠2 = 90° but in right △𝐶𝑃𝐸, 𝑚∠4 +𝑚∠1 = 90° . Thus, ∠1 ≅ ∠2 (both are complementary to ∠4), and right △𝐶𝑃𝐸 ≅ 𝑟𝑖𝑔ℎ𝑡 △𝐶𝐹𝐸, and 𝑃𝐸 = 𝐸𝐹. Similarly, 𝑚∠9 + 𝑚∠7+𝑚∠6 = 90° 𝑚∠9 + 𝑚∠7 = 𝑚∠5 . By substitution, 𝑚∠5 + 𝑚∠6 = 90°. However, in right △𝐶𝑃𝐷, 𝑚∠5 +𝑚∠7 = 90° Thus, ∠6 ≅ ∠7 (both are complementary to ∠5), and right ILM-FAN VA INNOVATSIYA ILMIY-AMALIY KONFERENSIYASI in-academy.uz/index.php/si △𝐶𝑃𝐷 ≅ 𝑟𝑖𝑔ℎ𝑡 △𝐶𝐺𝐷 and 𝐷𝑃 = 𝐷𝐺. Since 𝐷𝐸 = 𝐷𝑃 + 𝑃𝐸, we get 𝐷𝐸 = 𝐷𝐺 + 𝐸𝐹. Theorem-4. Prove that the sum of the measures of the perpendiculars from any point on a side of a rectangle to the diagonals is constant. (S-3) Proof. Let P be any point on side A̅̅̅B̅ of rectangle ABCD (Fig. S-5). P̅̅̅G̅ and P̅̅̅F̅ are perpendiculars to the diagonals. Draw 𝐴̅̅̅𝐽 perpendicular to 𝐷̅̅̅𝐵̅, and then 𝑃̅̅̅𝐻̅ perpendicular to 𝐴̅̅̅𝐽. Since 𝑃𝐻𝐽𝐹 is a rectangle (a quadrilateral with three right angles),we get 𝑃𝐹 = 𝐻𝐽. Since 𝑃̅̅̅𝐻̅̅ and 𝐵̅̅̅𝐷̅ are both perpendicular to 𝐴̅̅̅𝐽, 𝑃̅̅̅𝐻̅ is parallel to 𝐵̅̅̅𝐷̅. Thus, ∠𝐴𝑃𝐻 ≅ ∠𝐴𝐵𝐷. Since, 𝐴𝐸 = 𝐸𝐵, ∠𝐶𝐴𝐵 ≅ ∠𝐴𝐵𝐷. Thus, by transitivity, ∠𝐸𝐴𝑃 ≅ ∠𝐴𝑃𝐻; also in △𝐴𝑃𝐾, 𝐴𝐾 = 𝑃𝐾 . Since ∠𝐴𝐾𝐻 ≅ ∠𝑃𝐾𝐺 , right △𝐴𝐻𝐾 ≅ 𝑟𝑖𝑔ℎ𝑡 △𝑃𝐺𝐾 (S.A.A.). Hence, 𝐴𝐻 = 𝑃𝐺 and, by addition, 𝑃𝐹 + 𝑃𝐺 = 𝐻𝐽 + 𝐴𝐻