Prove the triangle center of gravity theorem. Example: It is known that △ABC, E and F are the midpoint between AB and AC. EC and FB to g.
Verification: EG= 1/2CG
Proof: cross e as EH∨BF and cross AC as H.
AE = BE，EH//BF
∴AH=HF= 1/2AF (parallel line segment ratio theorem)
AF = CF
∴HF= 1/2CF
∴HF:CF= 1/2
∫EH∨BF
∴EG:CG=HF:CF= 1/2
∴EG= 1/2CG
Method 2: Connect EF
Using triangle similarity
Verification: EG= 1/2CG proves EF= 1/2BC.
EF= 1/2BC can be proved by the midline, and EF= 1/2BC can be proved by the midline.
Prove the nature of judging the center of gravity of triangle;
△ In △ABC, three sides are A, B and C, point O is the center of gravity of the triangle, and AOA', Bob' and COC' are the median lines of sides A, B and C respectively. According to the nature of the center of gravity:
OA'= 1/3AA '
OB'= 1/3BB '
OC'= 1/3CC '
After o, a is higher than a, oh', ah, respectively.
It is known that OH'= 1/3AH.
Then, s △ BOC =1/2× oh 'a =1/2×1/3aha =1/3s △ ABC.
Similarly, it can be proved that S△AOC= 1/3S△ABC.
S△AOB= 1/3S△ABC
Therefore, S△BOC=S△AOC=S△AOB.
In triangle ABC, vector BO and vector BF*** lines, so you can set BO=xBF.
According to the triangle addition rule: vector AO=AB+BO.
=a+ xBF=a+ x(AF-AB)
= a+x(b/2-a)=( 1-x)a+(x/2)b
Vector CO and vector CD*** line, so we can set CO=yCD,
According to the triangle addition rule: vector ao = AC+co.
=b+ yCD=b+y(AD-AC)
= b+ y(a/2-b)=(y/2)a+( 1-y)b。
So the vector ao = (1-x) a+(x/2) b = (y/2) a+(1-y) b.
Then 1-x= y/2, x/2= 1-y,
X=2/3，y=2/3。
Vector BO=2/3BF, vector CO=2/3CD.
Namely bo: of = co: od = 2.
∴ vector ao = (y/2) a+(1-y) b =1/3a+1/3b.
The vector AE = ab+be = a+1/2bc = a+1/2 (AC-ab).
It is proved that the method of determining the center of gravity of triangle is known: 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 with AB at F. ..
Prove that f is the midpoint of AB. The center of gravity of the triangle.
Prove: According to the dovetail theorem, S△AOB=S△AOC, S△AOB=S△BOC,∴S△AOC=S△BOC, and then apply the dovetail theorem to get AF=BF, which proves the proposition.
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.
3. The sum of squares of the distances from the center of gravity to the three vertices of the triangle is the smallest.
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.
(y 1+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. Any connecting line between the center of gravity and the three vertices of the triangle divides the triangle area equally. Proof: It has just been proved that the intersection of three lines has been proved.
6. The center of gravity is the point where the distance product of a triangle to three sides is the largest.
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