An important theorem that all primary schools should master.
2. Projective Theorem (Euclid Theorem)
important
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.
important
4. The connecting lines of the two diagonal centers of the connecting lines of two sides of a quadrilateral intersect at one point.
A common problem in learning the middle line is not needed in the senior high school entrance examination, but in the junior high school competition.
5. The centers of gravity of two triangles formed by connecting the centers of the sides of a hexagon at intervals are coincident.
It makes no sense at all. You don't need to know the obvious conclusion after learning analytic geometry.
6. The perpendicular bisector of each side of a triangle intersects at one point.
important
7. Three vertical lines drawn from each vertex of a triangle to its opposite side intersect at one point.
important
8. Let the outer center of the triangle ABC be O and the vertical center be H. If a vertical line is drawn from O to BC and the vertical foot is not L, then AH=2OL.
You don't need a senior high school entrance examination, which is an obvious conclusion in the competition.
9. The outer center, vertical center and center of gravity of a triangle are on the same straight line.
A very important theorem in high school competition is called Euler line.
10, (nine-point circle or Euler circle or Fellbach circle) triangle, the center of three sides, drawn from each vertex to the vertical foot on the opposite side, connects the center of the vertical foot with the midpoint of each vertex, and these nine points are on the same circle.
Common theorems in high school competitions
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.
It will be used in high school competitions, but it is not often used.
12, Coolidge theorem: (A nine-point circle inscribed with a quadrilateral) There are four points on the circumference, any three of which are triangles, and the nine-point 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.
You don't need to master the topics of high school competitions.
13, the bisectors of the three internal 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)ss is half the circumference of the triangle.
important
14. A bisector of the inner corner of a triangle and the bisector of the outer corner at the other two vertices intersect at one point.
important
15, mean value theorem: (babs theorem) Let the midpoint of the side BC of the triangle ABC be p, then AB2+AC2=2(AP2+BP2).
Junior high school competition is necessary and important.
16, Stewart theorem: p If the side BC of the triangle ABC is divided into m:n, there is n×AB2+m×AC2=(m+n)AP2+mnm+nBC2.
High school competitions are necessary and important.
17, boromir and many theorems: 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.
The obvious conclusion does not need to be mastered.
18, Apollonius Theorem: Point P whose distance from two fixed points A and B is the constant ratio of 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.
High school competitions are necessary and important.
Ptolemy theorem: If the quadrilateral ABCD is inscribed in a circle, then AB×CD+AD×BC=AC.
Junior high school competition is necessary and important.
20. Taking the sides BC, CA and AB of any triangle ABC as the base, and making isosceles △BDC, △CEA and △AFB with the base angles outward of 30 degrees respectively, then △DEF is a regular triangle.
Learning complex numbers is an obvious conclusion and does not need to be mastered.
2 1, Elk's theorem 1: If △ABC and triangle △ are regular triangles, then the triangle formed by the barycentries of line segments AD, BE and CF is also regular triangles.
No need to master
22. Elkos Theorem 2: If △ABC, △DEF and △GHI are all regular triangles, then the triangle composed of the center of gravity △ADG, △BEH and △CFI is a regular triangle.
No need to master
23. Menelaus Theorem: Let the intersections of three sides BC, CA, AB or their extension lines △ABC and a straight line that does not pass through any of their vertices be p, q and r, respectively, then BPPC×CQQA×ARRB= 1.
Junior high school competition is necessary and important.
24. The inverse theorem of Menelaus theorem: (omitted)
Junior high school competition is necessary and important.
25. Application Theorem of Menelaus Theorem 1: Let the bisector A of △ABC intersect with the bisectors of Q and C, AB intersect with the bisectors of R and B, and CA intersects with Q, then the straight lines of P, Q and R are * * *.
No need to master
26. Menelaus Theorem Application Theorem 2: If the three vertices A, B and C of any △ABC are tangent lines of its circumscribed circle and intersect with the extension lines of BC, CA and AB at points P, Q and R respectively, then the three points P, Q and R are * * * lines.
No need to master
27. Seva Theorem: Let △ABC's three straight lines formed by the S-connecting surfaces of three vertices A, B and C that are not on the edge of a triangle or their extension lines intersect with the edges BC, CA, AB or their extension lines at points P, Q and R, then BPPC×CQQA×ARRB()= 1.
Junior high school competition is necessary and important.
28. The application theorem of Seva's theorem: Let the intersection of a straight line parallel to the side BC of △ABC and the sides AB and AC BE D and E respectively, and let Be and CD intersect with S, then AS will pass through the center M of the side BC.
No need to master
29. Inverse theorem of Seva theorem: (omitted)
Junior high school competition is necessary and important.
30. The Application Theorem of the Inverse Theorem of Seva Theorem 1: The three median lines of a triangle intersect at one point.
If this theorem is proved by Seva theorem, there is no geometric beauty, but it should be proved by the midline.
3 1, Application Theorem 2 of Inverse Theorem of Seva Theorem: If the inscribed circle of △ABC is tangent to points R, S and T respectively, then AR, BS and CT intersect at one point.
No need to master
32. simonson's Theorem: Take any point P on the circumscribed circle of △ABC as the vertical line to the three sides BC, CA, AB or their extension lines, and let its vertical foot be D, E, R, then D, E, R*** (this line is called simonson line).
Common theorems in junior high school competitions
33. The inverse theorem of simonson's theorem: (omitted)
Common theorems in junior high school competitions
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.
No need to master
35. The application theorem of Steiner's theorem: the symmetry point of a point P on the circumscribed circle of △ABC about the sides BC, CA and AB is on a straight line (parallel to Simpson line) with the vertical center H of △ABC. This straight line is called the mirror image of point P about △ABC.
No need to master
36. Blanchot and Tengxia Theorem: Let three points on the circumscribed circle of △ABC be P, Q and R, then the necessary and sufficient conditions for the intersection of P, Q and R about △ABC are: arc AP+ arc BQ+ arc CR=0(mod2∏).
No need to master
37. The protagonist and Tengxia Theorem Inference 1: Let p, q and r be three points on the circumscribed circle of △ABC. If the Simonson lines of p, q and r about △ABC intersect at one point, then the Simonson lines of a, b and c about △PQR also intersect at the same point as before.
No need to master
38. Inference 2 of Bolanger and Tengxia Theorem: In inference 1, the intersection of three Simpson lines is the midpoint of the line connecting the vertical centers of triangles made by any three points A, B, C, P, Q and R with the vertical centers of triangles made by other three points.
No need to master
39. Bolanger and 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 lines about △ABC at P, Q and R will intersect at one point.
No need to master
40. Inference 4 of Bolanger and Tengxia Theorem: Draw a vertical line from the vertex of △ABC to sides BC, CA and AB, let the vertical feet be D, E and F respectively, and let the midpoints of sides BC, CA and AB be L, M and N respectively, then six points of D, E, F, L, M and N are on the same circle, and then L, M,
No need to master
4 1, Theorem on Seymour Line1:△ The two endpoints P and Q of the circumscribed circle of ABC are perpendicular to each other with respect to Seymour Line of the triangle, and their intersection points are on the nine-point circle.
No need to master
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.
No need to master
43. Carnot Theorem: Through a point P of the circumscribed circle of △ABC, straight lines PD, PE and PF with the same direction and the same angle as the three sides BC, CA and AB of △ABC are introduced, and the intersections with the three sides are D, E and F respectively, so the three points of D, E and F are * * * lines.
No need to master
44. aubert's Theorem: Draw three parallel lines from the three vertices of △ABC, and their intersections with the circumscribed circle of △ABC are L, M and N respectively. If a point p is taken from the circumscribed circle of △ABC, the intersections of PL, PM and PN with BC, CA, AB or their extension lines are D, E and F respectively.
No need to master
45. Qing Palace Theorem: Let P and Q be two points of the circumscribed circle of △ABC, which are different from A, B and C respectively. The symmetry points of point P about BC, CA and AB are U, V and W respectively. At this time, the intersection of the inflections, and the edges BC, CA, and AB or their extension lines are D, E, and F, respectively, then D, F
No need to master
46. He takes the theorem: Let P and Q be a pair of antipodes about the circumscribed circle of △ABC, and the symmetry points of point P about BC, CA and AB are U, V and W respectively. At this time, if the intersections of flexion,, and edges BC, CA, AB or their extension lines are respectively ED, E and F, then D, E and F are three points. (anti-point: p and q are the radius OC of circle o and two points on its extension line, respectively. If OC2=OQ×OP, then the two points of P and Q are anti-points relative to the circle O)
No need to master
47. Langerhans Theorem: There is a point A1B1C1D14 on the same circle. Take any three points 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, then these four vertical feet are on the same straight line.
No need to master
48. Nine-point circle theorem: The midpoint of three sides of a triangle has three heights and three vertical feet of 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 an Euler circle and a Feuerbach circle.
It's already on it.
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.
No need to master
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.
No need to master
5 1, Cantor Theorem 2: If there are four points A, B, C and D and two points M and N on a circle, then the intersection points △BCD, △CDA, △DAB and △ABC of the two Simpsons of each of these four triangles are on the same straight line. This straight line is called the Cantor line about point M and point N of quadrilateral ABCD.
No need to master
52. Cantor Theorem 3: If there are four points A, B, C and D and three points M, N and L on a circle, then the Cantor line of quadrilateral ABCD at M and N, the Cantor line of quadrilateral ABCD at L and N intersect at one point. This point is called cantor point of m, n and l about quadrilateral ABCD.
No need to master
53. Cantor Theorem 4: If there are five points A, B, C, D and E and three points M, N and L on a circle, then the three points M, N and L are on a straight line with respect to each Cantor point in the quadrilateral BCDE, CDEA, DEAB and EABC. This line is called the Cantor line of M, N and L about pentagons A, B, C, D and E.
No need to master
54. Fairbach theorem: the nine-point circle of a triangle is tangent to the inscribed circle and the circumscribed circle.
No need to master
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.
This is one of the most beautiful and magical theorems in plane geometry, but I don't need to master it.
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.
Commonly used in high school competitions
57. Newton's Theorem 2: The midpoint, center and three-point * * * line of two diagonals of a circle circumscribed by a quadrilateral.
No need to master
58. Gilad Girard Desargues Theorem 1: There are two triangles △ABC and△ △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.
Occasionally used in high school games.
59. Gilad Girard Desargues Theorem 2: There are two triangles △ABC and △DEF in 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 Theorem: If vertices A and D, B and E, C and F of hexagonal ABCDEF tangent to a circle are connected, then these three lines are * * * points.
Occasionally used in high school games.
60. Pascal's theorem: a circle inscribed by the intersection (or extension line) of the opposite sides of a hexagon AB and DE, BC and EF, and CD and FA.
It is very important in high school competitions, generally called Pascal's theorem, which is a cone inscribed with a hexagon.