As shown in the graph, P is a point in the equilateral △ ABC, PA = 6, Pb = 8, PC = 10, if the point P 'is a point outside △ ABC, but △ p'ab is congruent with △ PAC The distance between point P and point P 'and the degree of ∠ APB

As shown in the graph, P is a point in the equilateral △ ABC, PA = 6, Pb = 8, PC = 10, if the point P 'is a point outside △ ABC, but △ p'ab is congruent with △ PAC The distance between point P and point P 'and the degree of ∠ APB

The ≌ P ≌ P ≌ P ≌ P ≌ P ? PAC ? PAC ? P ? P ≓ P ≓ P ? PAC ? P ≌ P ≌ P ≌ P ? PAC ? PAC ? PAC ≓ PAC ≓ PAC ? PAC e PAC ? P ? P ? B ? B + 60 = 15

As shown in the figure, P is a point in the equilateral triangle ABC, and PA = 6, Pb = 8, PC = 10. If △ PAC is rotated anticlockwise around point a to obtain △ P ′ AB, then the distance between point P and point P 'is () A. 4 B. 8 C. 10 D. 6

According to the property of rotation, P ′ a = PA = 6, ∠ BAP '= ∠ cap,
∵∠BAP=∠BAP，
Therefore, it can be concluded that: ∠ P ′ AP = ∠ BAC = 60 °,
ν Δ P ′ AP is an equilateral triangle,
∴P′P=PA=6．
Therefore, D

Given the three vertices a, B, C of triangle ABC and a point P in the plane, if PA + Pb + PC = AB, what is the position relationship between point P and triangle ABC

Vector PA + vector Pb + vector PC = vector ab
therefore
Vector PA + vector Pb + vector PC - vector AB = 0
Vector PA + vector Pb + vector PC + vector Ba = 0
Vector PA + vector PC + (vector Pb + vector BA) = 0
Vector PA + vector PC + vector PA = 0
So vector PC = - 2 vector PA
The equal point of P on AC

We know that the three vertices a, B, C of △ ABC and a point P in the plane satisfy the vector PA + vector Pb = vector PC. It is proved that P is outside the triangle!

PA + Pb = PC = > PA = pc-pb = CB, which means that vector PA and vector CB are parallel, then point P can only be outside the triangle

We know that the three vertices a, B, C of the triangle ABC and a point P in the plane satisfy the vector PA + vector Pb + vector PC = 0 if the real number λ

Known vector PA + vector Pb + vector PC = 0
Vector AB = vector Pb vector PA --- (1)
Vector AC = vector PC vector PA --- (2)
(1) + (2) = > vector AB + vector AC = vector Pb + vector PC-2 vector pa
λ vector AP = vector Pb + vector PC-2 vector pa
-λ vector PA = vector Pb + vector PC-2 vector pa
(2 - λ) vector PA = vector Pb + vector PC
(2 - λ) vector PA = - vector pa
(3 - λ) vector PA = 0
Since vector PA is not a zero vector, 3- λ =0, λ =3

Given that the three vertices a, B, C of the triangle ABC and a point P in the plane where the triangle ABC is located satisfy the vector PA + vector Pb + vector PC = vector AB, then where is the point P? The answer is on the trisection point of AC

∵ vector PA + vector Pb + vector PC = vector AB, the word "vector" is omitted below. Also, ab = pb-pa.

If the three vertices a, B, C of △ ABC and a point P in the plane of △ ABC are known, if PA+ PB+ PC＝ 0 if the real number λ satisfies AB+ AC＝λ AP, then the real number λ is equal to______ .

From the meaning of the title（
PB−
PA)+(
PC−
PA)＝−λ
PA；
∴(λ−2)
PA+
PB+
PC＝
Zero
∴λ=3．
So the answer is: 3

If PA = Pb = PC, then point O is () A. Be attached to one's heart B. Exocentrism C. Heart D. Center of gravity

∵ go through a point p outside the plane α where ⊥ ABC is located, and make Po ⊥ α, and the perpendicular foot is o,
Connect PA, Pb, PC. PA = Pb = PC,
∴OA=OB=OC，
The point O is the outer center of △ ABC
Therefore, B

Through a point p outside the plane of triangle ABC, make the vertical plane of Po, connect PA, Pb, PC, PA vertical Pb, Pb vertical PC, PC vertical PA, then o is triangle ABC what What kind of heart is it?

A: O is the vertical center of △ ABC
It is proved that: connect AO and extend BC to D, connect PD
∵ Po ⊥ plane ABC
BC in plane ABC
∴PO⊥BC
And ∵ PA ⊥ PC, PA ⊥ Pb
⊥ plane PBC
And ∵ BC is in plane PBC
∴PA⊥BC
ν BC ⊥ planar pad
∴BC⊥AD
That is, ad is higher than △ ABC
The other two Gao can prove the same

(1) if PA = Pb = PC, angle c = 90 degrees, then point O is ab side___ Why the midpoint

Because triangle ABC is a triangle with angle C as right angle, PA=PB=PC indicates that OA=DB=DC (knowledge of projection), from the midpoint on the hypotenuse to half of the hypotenuse, O is the midpoint of AB