Representation and detailed proof of each central vector of triangle in senior high school mathematics

Here are some of the contents I have compiled, hoping to help you:

Some conclusions: The following are all vectors.

1 if p is the center of gravity of △ABC, PA+PB+PC=0.

2 if p is PA of △ABC? PB=PB？ PC=PA？ PC (internal product)

3 If p is the inner APA+BP B+CPC of △ABC = 0 (ABC is trilateral).

4 If P is the outer center of △ABC |PA|? =|PB|？ =|PC|？

(AP means AP vector |AP| is its module)

5 AP=λ(AB/|AB|+AC/|AC|), λ∈[0, +∞) then the straight line AP passes through the heart of △ABC.

6 AP = λ (AB/| AB | COSB+AC/| AC | COSC), λ ∈ [0, +∞) passes through the center of gravity.

7 AP=λ(AB/|AB|sinB+AC/|AC|sinC)，λ∈[0，+∞)

Or AP=λ(AB+AC), λ∈[0, +∞) passes through the center of gravity.

8. If aOA=bOB+cOC, then 0 is the intersection of the centroid of ∠A and the bisector of ∠ B and C.

The following is the relevant proof of some conclusions.

O is the interior of a triangle if and only if aOA vector +bOB vector +cOC vector =0 vector.

Adequacy:

Known aOA vector +bOB vector +cOC vector =0 vector,

AcCOrding to vector addition, the co intersection AB is extended to d:

OA=OD+DA, OB=OD+DB, substituting for known:

a(OD+DA)+b(OD+DB) +cOC=0，

Because of the connection between OD and OC***, OD=kOC can be set.

The above formula can be replaced by a vector of (ka+kb+c) OC+( aDA+bDB)=0.

Vector DA and DB*** lines, vector OC and vectors DA and DB are not * * * lines,

So there can only be: ka+kb+c=0, aDA+bDB=0 vectors,

According to the vector of aDA+bDB=0, the ratio of the length of DA to DB is b/a,

So CD is the bisector of ∠ACB, which proves that the other two are also angular bisectors.

Necessity:

As we all know, O is the center of a triangle.

Let BO and AC intersect at e, and CO and AB intersect at f,

O is the heart

∴b/a=AF/BF,c/a=AE/CE

The parallel lines passing through a as CO intersect with the extension line of B0 at n, and the parallel lines passing through a as BO intersect with the extension line of c0 at m,

So the quadrilateral Oman is a parallelogram.

According to the parallelogram law, it is concluded that

Vector OA

= vector OM+ vector ON

=(OM/CO)* vector CO+(ON/BO)* vector BO

=(AE/CE)* Vector CO+(AF/BF)* Vector BO

=(c/a)* vector CO+(b/a)* vector BO∴a* vector OA=b* vector BO+c* vector CO.

∴a* vector OA+b* vector OB+c* vector OC= vector 0.

It is known that △ABC is an oblique triangle, O is a fixed point on the plane where △ABC lies, and the moving point P satisfies the vector OP = OA+ENTER {(AB/| AB | 2 * SIN2B)+AC/(| AC | 2 * SIN2C)}.

Find the vertical center of the locus of point P passing through the triangle.

OP = OA+enter {(ab/| ab | 2 * sin2b)+AC/(| AC | 2 * sin2c)}，

OP-OA = enter {(ab/| ab | 2 * sin2b)+AC/(| AC | 2 * sin2c)}，

AP = enter {(ab/| ab | 2 * sin2b)+AC/(| AC | 2 * sin2c)}，

AP？ BC= enter {(AB？ BC/| AB | 2 * SIN2B)+AC? BC/(| AC | 2 * sin2c)},

AP？ BC= enter {|AB|？ | BC | cos( 180-b)/(|ab|^2*sin2b)+| AC |？ | BCE | COSC/(| AC | 2 * SIN2C)},

AP？ BC= enter {-|AB|？ | BC | cos b/(|ab|^2*2sinb cos b)+| AC |？ |BC| cosC/(|AC|^2*2sinC cosC)}，

AP？ BC = enter {-| BC |/(| AB | * 2 sinb)+| BC |/(| AC | * 2 sinc)}，

According to sine theorem: |AB|/sinC=|AC|/ sinB, so |AB|*sinB=|AC|*sinC.

∴-|bc|/(| ab | * 2 sinb)+| BC |/(| AC | * 2 sinc)= 0，

Is that AP? BC=0，

The locus of point P passes through the vertical center of the triangle.