In a triangle, the intersection of bisectors of three angles is the center of the inscribed circle of the triangle, which is called the heart of the triangle.

The principle that the center is the intersection of the bisector of a triangle: make two tangents of the circle through a point outside the circle, and the connecting line between this point and the center of the circle bisects the included angle of these two tangents (principle: the distance from the point on the bisector of the angle is equal to both sides of the angle).

Interior Theorem: The bisectors of three interior angles of a triangle intersect at one point. This point is called the heart of a triangle.

Note that the distance from the inner heart to the three sides is equal (the radius of the inscribed circle), and the inner heart theorem is actually easy to prove.

If the three sides are l 1, l2, l3, and the perimeter is p, then the coordinates of the center of gravity of the heart are (l 1/p, l2/p, l3/p).

The distance from the center to the side of a right triangle is equal to the sum of two right angles minus half the difference of the hypotenuse.

The projection of the center of a triangle consisting of a point and two focal points on any branch of a hyperbola on the real axis is the vertex of the corresponding branch.

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The nature of the heart of a triangle

Let the inscribed circle of △ABC be ☉I(r), and the opposite sides of angles A, B and C are A, B and C respectively, and p = (a+b+c)/2.

1, the three bisectors of the triangle intersect at one point, which is the heart of the triangle.

2. The distance from the center of the triangle to the three sides is equal, which is equal to the radius r of the inscribed circle.

3. r = Standard & Poor's.

4. In Rt△ABC, ∠ c = 90, r = (a+b-c)/2.

5、∠BIC=90 +A/2。

6. The necessary and sufficient conditions for point O to be any point on the plane ABC and point I to be △ABC are as follows:

A (vector OA)+b (vector OB)+c (vector OC)= vector 0.

7. The necessary and sufficient conditions for point O to be any point on the plane ABC and point I to be △ABC are as follows:

Vector OI=[a (vector OA)+b (vector OB)+c (vector oc)]/(a+b+c).

8. In △ABC, A(x 1, y 1), B(x2, y2) and C(x3, y3), then the coordinates of the internal I of △ABC are:

(ax 1/(a+b+c)+bx2/(a+b+c)+cx3/(a+b+c)，ay 1/(a+b+c)+by2/(a+b+c)+cy3/(a+b+c))。

9. In (euler theorem) ⊿ABC, where R and R are the radii of circumscribed circle and inscribed circle respectively, and O and I are the outer center and inner center respectively, then Oi 2 = R 2-2rr.

10, (the bisector of the inner corner is divided into three sides)

△ABC, where 0 is the center, and the bisectors of internal angles of ∠A, ∠B and ∠C intersect with BC, AC and AB in Q, P and R respectively, then BQ/QC = C/B, CP/PA = A/C, and Br/Ra = A/B. 。

External center definition

The center of the circumscribed circle of a triangle is called the outer center of the triangle.

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The nature of the outer center of triangle

Let the circumscribed circle of ⊿ABC be G (r), and the opposite sides of angles A, B and C are A, B and C respectively, and P = (A+B+C)/2.

1. The intersection of perpendicular bisector of three sides of a triangle is the center of the circumscribed circle of the triangle.

2. The outer center of the acute triangle is in the triangle; The outer center of an obtuse triangle is outside the triangle; The outer center of a right triangle is on the hypotenuse and coincides with the midpoint of the hypotenuse.

3、GA=GB=GC=R。

3.∠BGC=2∠A or ∠ BGC = 2 (180-∠ A).

4、R=abc/4S⊿ABC.

5. If the G point is a point on the plane ABC, then the necessary and sufficient conditions for the G point to be ⊿ABC's outer center are:

(Vector GA+ Vector GB) Vector AB= (Vector GB+ Vector GC) Vector BC= (Vector GC+ Vector GA) Vector CA= Vector 0.

6. If point G is a point on the plane ABC and point P is an arbitrary point on the plane ABC, then the necessary and sufficient conditions for point G to be the outer center of ⊿ABC's are as follows:

Vector PG=((tanB+tanC) vector PA+(tanC+tanA) vector PB+(tanA+tanB) vector PC)/2(tanA+tanB+tanC).

7. If point G is a point on the plane ABC and point P is an arbitrary point on the plane ABC, then the necessary and sufficient conditions for point G to be the outer center of ⊿ABC's are as follows:

Vector PG=(cosA/2sinBsinC) vector PA+(cosB/2 sin Sina) vector Pb+(COSC/2 sinsine nb) vector PC.

8. Let d 1, d2 and d3 be the point multiplication of the vectors connecting the three vertices and the other two vertices of a triangle respectively. c 1=d2d3，c2=d 1d3，C3 = d 1 D2； c=c 1+c2+c3 .

Gravity center coordinates: ((c2+c3)/2c, (c 1+c3)/2c, (c 1+c2)/2c).

9. The distances from the outer center to the three vertices are equal.

center of gravity,core,barycenter,main point

The center of gravity is the intersection of three sides of a triangle, and the intersection of three lines can be proved by dovetail theorem. The proof process is a special case of Seva theorem.

It is known that in △ABC, D is the midpoint of BC, E is the midpoint of AC, AD and BE intersect at O, and the extension line of CO intersects AB at F. ..

Prove that f is the midpoint of AB.

Prove: According to the dovetail theorem, S△AOB=S△AOC, S△AOB=S△BOC, ∴S△AOC=S△BO.

C, and then smear from the middle point.

AF=BF, proposition proof.

Some properties of the center of gravity and their proof methods;

1, and the ratio of the distance from the center of gravity to the vertex to the distance from the center of gravity to the midpoint of the opposite side is 2: 1.

2. The areas of the three triangles formed by the center of gravity and the three vertices of the triangle are equal.

Proof method:

▲ In ▲ABC, the three sides are A, B and C, the point O is the center of gravity of the triangle, and the median lines of AOA 1, BOB 1 and COC 1 are A, B and C respectively. According to the nature of the center of gravity, OA 1 = 1/3A65438. OC1=1/3cc1crosses O, where A is h 1 on the side of A, H represents H1=13h, and S (▲ boc) =/kloc-0. It can also be proved that S(▲AOC)= 1/3S(▲ABC), S(▲AOB)= 1/3S(▲ABC) Therefore, S(▲BOC)=S(▲AOC)=S(▲AOB).

3. The sum of the distances from the center of gravity to the three vertices of the triangle is the smallest.

Proof method:

Let three vertices of a triangle be (x 1, y 1), (x2, y2), and any point on the plane of (x3, y3) be (x, y), then the sum of the distances from that point to the three vertices is: (x1-x) 2+(y). +x 1^2+x2^2+x3^2+y 1^2+y2^2+y3^2=3(x- 1/3*(x 1+x2+x3))^2+3(y- 1/3(y 1+y2+y3))^2+x 1^2+x2^2+x3^2+y / kloc-0/^2+y2^2+y3^2- 1/3(x 1+x2+x3)^2- 1/3(y 1+y2+y3)^2

Obviously, when X = (X 1+X2+X3)/3 and Y = (Y 1+Y2+Y3)/3 (barycentric coordinates), the above formula gives x12+x2 2+x3 2+y/.

4. In the plane rectangular coordinate system, the coordinate of the center of gravity is the arithmetic average of the vertex coordinates, that is, its coordinate is ((x 1+x2+x3)/3, (y1+y2+y3)/3); Spatial rectangular coordinate system-abscissa: (X 1+X2+X3)/3 ordinate: (Y 1+Y2+Y3)/3 ordinate: (z 1+z2+z3)/3.

5. The point where the product of the distance from a triangle to three sides is the largest.

Chongxin

The three midlines must intersect, and the intersection position is really strange.

The intersection is named "center of gravity", and the nature of the center of gravity should be clear.

In the center of gravity segmentation, you can hear the ratio of line segment to several segments;

The ratio of length to length is two to one, so we should use it flexibly and master it well.

Definition of vertical center

The heights of three sides of a triangle intersect at a point, which is called the vertical center of the triangle.

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Properties of vertical center of triangle

Let ⊿ABC have three heights: AD, BE and CF, where D, E and F are vertical feet, the vertical center is H, and the opposite sides of angles A, B and C are A, B and C, respectively, and p = (a+b+c)/2.

1, the vertical center of the acute triangle is within the triangle; The vertical center of a right triangle is at the right vertex; The vertical center of an obtuse triangle is outside the triangle.

2. The vertical center of a triangle is the center of its vertical base triangle; In other words, the heart of a triangle is the center of the triangle next to it;

3. The symmetry points of the vertical center H about three sides are all on the circumscribed circle of △ABC.

4. There are six groups of four-point * * * circles and three groups (four in each group) of similar right-angled triangles in △ ABC, Ah HD = BH He = CH HF.

5. Any of the four points H, A, B and C is the vertical center of a triangle with the other three points as its vertices (such four points are called vertical center groups).

6. The circumscribed circles of △ ABC, △ABH, △BCH and △ACH are all equal circles.

7. In a non-right triangle, if the straight line passing through H intersects AB and AC in P and Q respectively, AB/AP Tanb+AC/AQ Tanc = Tana+Tanb+Tanc.

8. The distance from any vertex of a triangle to the vertical center is equal to twice the distance from the outer center to the opposite side.

9. Let O and H be the external center and vertical center of △ABC respectively, then ∠BAO=∠HAC, ∠ABH=∠OBC, ∠BCO=∠HCA.

10, the sum of the distances from the vertical center of an acute triangle to the three vertices is equal to twice the sum of the radii of its inscribed circle and circumscribed circle.

1 1, and the vertical center of the acute triangle is the center of the vertical triangle; Among the inscribed triangles of acute triangle (the vertex is on the side of the original triangle), the perimeter of vertical triangle is the shortest.

12、

Siemsen Theorem (siemsen Line)

An important condition for drawing a perpendicular from a point to three sides of a triangle is that the point falls on the circumscribed circle of the triangle.

I hope it helps you.