Let p be a point in the plane of triangle ABC, and vector AP + 2 vector BP + 3 vector CP = vector 0, If vector AB = vector a, vector AC = vector B. (1) use vector a and vector B to represent vector AP and vector ad, (2) fill in the blank s triangle according to the above results PAB:S triangle PBC:S Triangle PAC = --- - -:----

Let p be a point in the plane of triangle ABC, and vector AP + 2 vector BP + 3 vector CP = vector 0, If vector AB = vector a, vector AC = vector B. (1) use vector a and vector B to represent vector AP and vector ad, (2) fill in the blank s triangle according to the above results PAB:S triangle PBC:S Triangle PAC = --- - -:----

(1) Vector AP + 2 vector BP + 3 vector CP = vector 0
According to the subtraction of vector, we can know that vector AP + 2 vector (ap-ab) + 3 vector (ap-ac) = vector 0
That is, 6ap-2ab-3ac = 0,
The vector AP = 1 / 3AB + 1 / 2Ac = 1 / 3A + 1 / 2B
Vector ad is collinear with vector AP, so there is a unique real number m such that
Vector ad = m vector AP = m (1 / 3A + 1 / 2b) = 1 / 3mA + 1 / 2MB
Vector BD is collinear with vector BC, so there is a unique real number n such that
Vector BD = n vector BC = n (ac-ab) = n (B-A),
The vector ad = AB + BD = a + n (B-A) = (1-N) a + Nb
In conclusion, vector ad = 1 / 3mA + 1 / 2MB = (1-N) a + Nb
So 1 / 3M = (1-N), 1 / 2m = n,
5 / N = 5
So the vector ad = 1 / 3mA + 1 / 2MB = 2 / 5A + 3 / 5B
(2) From (1) we know that: vector BD = 3 / 5 vector BC, vector ad = 6 / 5 vector AP
Therefore, the height of △ PBD is the same as that of △ PDC, and the ratio of BD to DC at the bottom is 3:2,
So the area ratio is 3:2,
If the area of △ PBD is 3S, then the area of △ PDC is 2S
The height of △ PBD is the same as that of △ ABP, and the ratio of PD to AP at the bottom is 1:5,
So the area of △ ABP is 15s
The height of △ PDC is the same as that of △ ACP, and the ratio of PD to AP at the bottom is 1:5,
So the area of △ ABP is 10s
The s triangle PAB:S triangle PBC:S Triangle PAC = 15s: 5S: 10s = 3:1:2

Given that D is the midpoint of the ABC edge BC of the triangle, point P satisfies the vector PA + BP + CP = 0, AP = RPD, then the value of real number r

Point P satisfies: PA + BP + CP = 0
PA=PB＋PC
Connect AD and extend to Q such that: QD = Da
Then:
Vector QA = QB + QC
That is to say, point q is point P
If AP = PD, then:
r=－2

Find a point P in the triangle ABC to minimize the vector AP + vector BP + vector CP

It is proved that the coordinates of points a, B and C are (x1, Y1) (X2, Y2) (X3, Y3) respectively. From the formula of barycenter coordinates, P [(x1 + x2 + x3) / 3, (Y1 + Y2 + Y3) / 3] can be obtained

Let p be a point in the triangle ABC, and 3 vector AP + 4 vector BP + 5 vector CP = vector o, extend AP and intersect BC at point D. if vector AB = vector a, vector AC = vector B, vector a, B denote vector AP, ad

In this paper, we have studied the effect of X-1 / 3 = 1 / (3K) and Y-5 / 12 = 5, which is: 3aP + 4 (ap-ab) + 5 (ap-ac) = 12ap-4ab-5ac = 0, that is: AP = 4AB / 12 + 5AC / 12 = (4a + 5b) / 12: AP = KPD, then: PD = (4AB + 5AC) / (12K): ad = XAB + YAC, then: x + y = 1 then: PD = ad-ap = (x-4 / 12) AB + (Y-5 / 12) AC

Find a point in △ ABC to minimize (vector AP) ^ 2 + (vector BP) ^ 2 + (vector CP) ^ 2. What special point is point P?

The point P is the center of gravity of △ ABC. It is obvious that there are: vector AP = vector CP - vector Ca, vector BP = vector CP - vector CB, ν (vector AP) ^ 2 + (vector BP) ^ 2 + (vector CP - vector CA) ^ 2 + (vector CP - vector CB) ^ 2 + (vector CP) ^ 2 = (vector CP) ^ 2

The geometric proof of the second grade of junior high school shows that the angle a = 60 degrees, the angle c = 40 degrees, P, q are on BC and AC, and AP and BQ are the bisectors of angle A and angle B, and BQ + AQ = AB + BP

First, the angle QBC = 40 = angle QCB, so BQ = CQ, BQ + AQ = CQ + AQ = AC
Secondly, extend AB to D to make ad = AC, connect CD and PD,
Obviously, ACD is an equilateral triangle and AP is the vertical bisector of CD,
So PC = PD, angle PDC = angle PCD = 60-40 = 20 degrees
Thus the angle BDP = 60-20 = 40 degrees, the angle BPD = angle PCD + angle PDC = 20 + 20 = 40 degrees
BP = BD
Therefore, AB + BP = AB + BD = ad = AC = BQ + aq
Conclusion

As shown in the figure, in △ ABC, ∠ BAC = 60 °, ACB = 40 °, P and Q are respectively on BC and Ca, and AP and BQ are angular bisectors of ∠ BAC and ∠ ABC （1）BQ=CQ；    （2）BQ+AQ=AB+BP．

It is proved that: (1) ∵ BQ is the angular bisector of ∵ ABC,
∴∠QBC=1
2∠ABC．
 ABC + ∠ ACB + ∠ BAC = 180 ° and ∠ BAC = 60 ° and ∠ ACB = 40 °,
∴∠ABC=80°，
∴∠QBC=1
2×80°=40°，
∴∠QBC=∠C，
∴BQ=CQ；
(2) Extend AB to m so that BM = BP and connect MP
∴∠M=∠BPM，
∵ in ∵ ABC, ∵ BAC = 60 ° and ∵ C = 40 °,
∴∠ABC=80°，
∵ BQ bisection ∵ ABC,
∴∠QBC=40°=∠C，
∴BQ=CQ，
∵∠ABC=∠M+∠BPM，
∴∠M=∠BPM=40°=∠C，
∵ AP bisection ∵ BAC,
∴∠MAP=∠CAP，
In △ amp and △ ACP,
A kind of
∠M＝∠C
∠MAP＝∠CAP
AP＝AP
∴△AMP≌△ACP，
∴AM=AC，
∵AM=AB+BM=AB+BP，AC=AQ+QC=AQ+BQ，
∴AB+BP=AQ+BQ．

As shown in the figure, in the triangle ABC, ab = AC, ∠ BAC = 90 ° and take the point P on ab. take the point Q on the extension line of Ca so that AP = AQ, and the edge CP and BQ intersect with the point s. It is proved that △ cap is all equal to △ BAQ

AB = AC,
∠BAQ=∠CAP=90°,
AQ=AP,
∴△BAQ≌△CAP（SAS）
The proof is over

As shown in the figure, if P and Q are two points on the edge BC of △ ABC, and BP = PQ = QC = AP = AQ, then the size of ∠ ABC is equal to______ Degree

∵PQ=AP=AQ，
△ Apq is an equilateral triangle,
∴∠APQ=60°，
And ∵ AP = BP,
∴∠ABC=∠BAP，
∵∠APQ=∠ABC+∠BAP，
Therefore, the size of ∠ ABC is equal to 30 °
So the answer is 30 degrees

Let P and Q be two points on the edge BC of △ ABC, and BP = PQ = QC = AP = AQ, find the degree of ∠ BAC

Because AP = PQ = AQ, the triangle Apq is an equilateral triangle, each angle is 60 degrees
Because AP = Pb, ABP is an isosceles triangle, because angle Apq = 60, so angle APB = 120, so angle BAP = 30,
Similarly, the angle QAC = 30, so the degree of ∠ BAC = 60 + 30 + 30 = 120 degrees