Plane geometry is the key and difficult point in junior middle school mathematics. Students must master the theorem of plane geometry if they want to learn the plane geometry problems in junior middle school well. Let me introduce you to the theorem of plane geometry. I hope I can help you.

1. Pythagorean Theorem (Pythagorean Theorem)

2. Projective Theorem (Euclid Theorem)

3. The three median lines of a triangle intersect at a point, and each median line is divided into two parts of 2: 1 by this point.

4. The connecting lines of the two diagonal centers of the connecting lines of two sides of a quadrilateral intersect at one point.

5. The centers of gravity of two triangles formed by connecting the centers of the sides of a hexagon at intervals are coincident.

6. The perpendicular bisector of each side of a triangle intersects at one point.

7. The three high lines of a triangle intersect at one point.

8. Let the outer center of triangle ABC be O, the vertical center be H, draw a vertical line from O to BC, and the vertical foot be L, then AH=2OL.

9. The outer center, vertical center and center of gravity of a triangle are on the same straight line (Euler line).

10. In a triangle (nine-point circle or Euler circle or Fellbach circle), the centers of three sides, the vertical foot of the vertical line drawn from each vertex to its opposite side, and the midpoint of the line connecting the vertical center and each vertex are all on the same circle.

1 1. euler theorem: The outer center, center of gravity, center of nine o'clock and vertical center of a triangle are located on the same straight line (Euler line) in turn.

12. Coolidge's maximum theorem: (nine-point circle inscribed with quadrilateral)

There are four points on the circumference, any three of which are triangles, and the nine centers of these four triangles are on the same circumference. We call the circle passing through these four nine-point centers a nine-point circle inscribed with a quadrilateral.

13. The bisectors of the three inner angles of the (inner) triangle intersect at one point, and the radius formula of the inscribed circle is: r=(s-a)(s-b)(s-c)s, where s is half the circumference of the triangle.

14. The inner bisector and the outer bisector of the triangle intersect at one point at the other two vertices.

15. midline theorem: (babs theorem) let the midpoint of the side BC of the triangle ABC be p, then AB2+AC2=2(AP2+BP2).

16. Stewart Theorem: p Divide the side BC of the triangle ABC into m:n, then there is n? AB2+m？ AC2=(m+n)AP2+mnm+nBC2

17. Boromir Theorem: When the diagonals of the quadrilateral ABCD inscribed in a circle are perpendicular to each other, the straight line connecting the midpoint M of AB and the diagonal intersection E is perpendicular to CD.

18. Apollonius Theorem: Point P whose distance to two fixed points A and B is constant ratio m:n (the value is not 1) is located on a fixed circle with the inner bisector C and the outer bisector D which divide the line segment AB into m:n at both ends of the diameter.

Ptolemy theorem: if a quadrilateral ABCD is inscribed in a circle, then there is AB? CD+AD？ BC=AC？ Bachelor of science

20. Stand on the side of BC. Take about AB of any triangle ABC as the base and make isosceles △BDC. △CEA。 △ AFB with a base angle of 30 degrees outward, then △DEF is a regular triangle.

2 1. Elkos Theorem 1: If both △ABC and △DEF are regular triangles, the triangle formed by the center of line segment AD. BE.CF is also a regular triangle.

22. Elkos Theorem 2: If △ABC. △DEF。 △GHI is a regular triangle, then the triangle formed by the center of gravity △ADG of the triangle. △BEH。 △CFI is a regular triangle.

23. Menelaus theorem: If the intersection of three sides BC. CA.AB or its extension line △ABC and a straight line that does not pass through any of their vertices are P.Q.R Will there be BPPC? CQQA？ ARRB= 1

24. The inverse theorem of Menelaus theorem: (omitted)

25. application theorem of menelios theorem 1: Set △ABC? Is the intersection point CA of the bisector of the outer angle of A in Q? The intersection point CA of the bisector of C and AB at R and B is in Q, then the P.Q.R three-point line * * *.

26. Application Theorem 2 of Menelaus Theorem: If the three vertices A, B and C of any △A.B.C are tangents of its circumscribed circle and intersect with the extension lines of BC, CA and AB at points P, Q and R, then the three points P, Q and R are * * lines.

27. Seva's Theorem: Let three straight lines or their extension lines formed by point S of the connection surface of three vertices A, B and C which are not on the edge of a triangle intersect with edges BC, CA, AB or their extension lines at points P, Q and R respectively, then BPPC? CQQA？ ARRB()= 1。

28. The application theorem of Seva theorem: Let the intersection of a straight line be parallel to the side BC of △ABC and AB. AC on both sides is D.E, let BE and CD intersect s, then AS must pass through the center m of BC side.

29. Inverse theorem of Seva theorem: (omitted)

30. The Application Theorem of the Inverse Theorem of Seva Theorem 1: The three median lines of a triangle intersect at one point.

3 1. Application Theorem 2 of the Inverse Theorem of Seva Theorem: Let △ABC and the inscribed circle of the edge BC. About AB is tangent to the point R.S.T, and then tangent to AR. BS.CT intersects at one point.

32. simonson's Theorem: From any point P on the circumscribed circle of △ABC to trilateral BC. CA.AB or its extension line is a vertical line, and if its vertical foot is D.E.R, it is a D.E.R*** line (this line is called simonson line).

33. The inverse theorem of Seymour's theorem: (omitted)

34. Steiner theorem: Let the vertical center of △ABC be H, which circumscribes any point p of the circle. At this time, the Simpson line of point p about △ABC passes through the center of the line segment pH.

35. The application theorem of Steiner's theorem: the symmetric point of point P on the circumscribed circle of △ABC of side BC. The vertical center h of CA.AB and △ABC is on the same straight line (parallel to Seymour line). This straight line is called the mirror image of point P about △ABC.

36. Bolanjetenxia Theorem: Let three points on the circumscribed circle of △ABC be P.Q.R, then the necessary and sufficient conditions for the intersection of P.Q.R and △ABC at one point are: arc AP+ arc BQ+ arc CR=0(mod2? ).

37. inference of bolanjietenxia theorem 1: let P.Q.R be three points on the circumscribed circle of △ABC. If the Seymour line of P.Q.R about △ABC intersects at one point, then the Seymour line of A.B.C about △PQR intersects at the same point as before.

38. Inference 2 of Bolanjetenxia Theorem: In inference 1, the intersection of three Seymour lines is the midpoint of the connecting line between the vertical center of the triangle made by A.B.C.P.Q.R at six points and the vertical center of the triangle made by the other three points.

39. Bolanger. Tengxia Theorem Inference 3: Investigate the Simpson line of point P on the circumscribed circle of △ABC. If QR is perpendicular to this Simpson line, the Simpson line about △ABC at three points of P.Q.R will intersect at one point.

40. inference 4 of bolanjetenxia theorem: draw a vertical line from the vertex of △ABC to the side of BC. CA.AB, let the vertical foot be D.E.F and the side length be the midpoint of BC. CA.AB is L.M.N, then six points of D.E.F.L.M.N are on the same circle, and then l.m.n.

4 1. Theorem on Seymour Line 1:△ABC The two endpoints of the circumscribed circle are perpendicular to each other with respect to the Seymour line of the triangle, and their intersection points are on the nine-point circle.

42. Theorem 2 on Simpson Line (Peace Theorem): There are four points on a circle, any three of which are triangles, and then the remaining points are Simpson lines about triangles, and these Simpson lines intersect at one point.

43. Carnot's Theorem: A point P passing through the circumscribed circle of △ABC, a straight line PD. Introduce PE.PF with the same direction and equal angle as the three sides of △ABC, and the intersection point with the three sides is D.E.F, which is the D.E.F three-point * * line.

44. aubert's Theorem: Draw three parallel lines from the three vertices of △ABC, and their intersection with the circumscribed circle of △ABC is L.M.N If a point P is taken from the circumscribed circle of △ABC, then the intersection of PL. Afternoon, PN and BC. CA.AB or their extension lines are D.E.F, and then d.e.f.

45. qinggong theorem: let P.Q be two points on the circumscribed circle of △ABC, which are different from A.B.C, and point p is the symmetrical point of trilateral BC. At this time, CA.AB is the intersection of U.V.W and flexion. QV。 QW and BC. CA.AB or their extension lines are D.E.F, then d.

46. He took the theorem: Let P.Q be a pair of antipodes about the circumscribed circle of △ABC, and point P on the three sides of BC is the symmetrical point. CA.AB is U.V.W At this time, if the intersection point bends. QV。 QW and BC. CA.AB or their extension lines are ED. E.f. respectively, then d.e. (anti-point: P.Q is the radius OC of the circle o and two points on its extension line respectively. If OC2=OQ? OP says that the two points of P.Q are about the anti-point of circle o)

47. Langerhans Theorem: There is a point A1B1C1D14 on the same circle. Take any three of them as triangles, take a point P on the circumference, make a point P about Seymour lines of these four triangles, and then draw a vertical line from P to these four Seymour lines, so the four feet are on the same straight line.

48. Nine-point circle theorem: the midpoint of three sides of a triangle, the vertical foot of three heights and three Euler points [the midpoint of three line segments obtained by connecting the vertex of the triangle with the vertical center] a nine-point * * * circle [usually called a nine-point circle], or Euler circle, Feuerbach circle.

49. There are n points on a circle, and the perpendicular lines drawn from the center of gravity of any n- 1 point to the tangents of other points of the circle intersect at one point.

50. Cantor Theorem 1: There are n points on a circle, and the perpendicular line drawn from the center of gravity of any n-2 points to the other two points is * * * points.

5 1. cantor theorem 2: if there are four points of A.B.C.D and two points of M.N on a circle, then the intersection of two Simpsons of m and n about each of the four triangles is △BCD. △CDA。 △DAB。 △ABC is on the same straight line. This straight line is called cantor line between two points of M.N about quadrilateral ABCD.

52. Cantor Theorem 3: If there are four points of A.B.C.D and three points of M.N.L. on a circle, Cantor line of M.N. about quadrilateral ABCD, Cantor line of L.N. about quadrilateral ABCD and Cantor line of M.L. about quadrilateral ABCD intersect at one point. This point is called the Cantor point of the M.N.L three points about the quadrilateral ABCD.

53. Cantor Theorem 4: If there are five points A.B.C.D.E and three points M.N.L on a circle, then these three points M.N.L are on a straight line with respect to each Cantor point in the quadrilateral BCDE.CDEA.DEAB.EABC. This straight line is called M.N.L about Pentagon A.B.C.

54. Fairbach theorem: the nine-point circle of a triangle is tangent to the inscribed circle and the circumscribed circle.

55. Morley Theorem: If three internal angles of a triangle are divided into three equal parts and two bisectors near one side get an intersection point, then such three intersections can form a regular triangle. This triangle is usually called Molly's regular triangle.

56. Newton's theorem 1: the midpoint of the line segment connected by the intersection of the extension lines of two opposite sides of a quadrilateral and the midpoint of two diagonal lines, three * * * lines. This straight line is called Newton line of this quadrilateral.

57. Newton's Theorem 2: The midpoint, center and three-point * * * line of two diagonals of a circle circumscribed by a quadrilateral.

58. Gilad Girard Desargues Theorem 1: There are two triangles △ABC. △ DEF on the plane. Let the connecting lines of their corresponding vertices (a and D.B and E.C and f) intersect at one point. At this time, if the corresponding edges or their extension lines intersect, the three intersection points are * * * lines.

59. Gilad Girard Desargues Theorem 2: There are two triangles △ABC. △ DEF on different planes. Let the connecting lines of their corresponding vertices (a and D.B and E.C and f) intersect at one point. At this time, if the corresponding edges or their extension lines intersect, the three intersection points are * * * lines.

60. Bryansson's theorem: connecting the opposite vertices A and D.B and E.C and F of hexagonal ABCDEF tangent to the circle, these three lines are * * * points.

Pascal's theorem: A circle is inscribed in the intersection (or extension line) of the opposite side AB of the hexagon ABCDEF and DE. BC and EF. CD and FA.

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