Given that point P is in the plane of triangle ABC, vector PA * Pb = Pb * PC = PC * PA, how to prove that P is the perpendicular of triangle?

Given that point P is in the plane of triangle ABC, vector PA * Pb = Pb * PC = PC * PA, how to prove that P is the perpendicular of triangle?

∵ vector PA · vector Pb = vector PC · vector PA, ∵ vector PA · vector Pb - vector PA · vector PC = 0,
The vector PA · (vector Pb vector PC) = 0, the vector PA · vector CB = 0, the vector PA ⊥ vector CB,
∴PA⊥CB.
Similarly, from the vector PA · vector Pb = vector Pb · vector PC, we can get: Pb ⊥ ca
From PA ⊥ CB and Pb ⊥ Ca, it is concluded that point P is the intersection of the heights of CB and Ca in △ ABC, and point P is the perpendicular of △ ABC

In the triangle ABC, AB is the longest side and P is a point in the triangle. It is proved that PA + Pb > PC

PA+PB>AB
We hereby certify that PC must be smaller than at least one of AC and BC
(counter evidence) suppose PC > AC and PC > BC
Taking C as the center of the circle and the length of PC as the radius, the locus of the moving point P, that is, the arc, falls outside △ ABC, which contradicts that P is a point inside △ ABC
Therefore, the hypothesis does not hold
｜ PC ＜ AC or PC ＜ BC
AC < AB and BC < ab
∴PC＜AB ∴PA+PB＞PC

The triangle ABC is an equilateral triangle P is a point outside the triangle PA = Pb + PC prove that P, a, B, C are all round

Through point C and P, we can make equilateral triangle DCP, and connect BD, then ∠ DCP = ∠ ACB = 60 °∠ ACP = ∠ BCD. Then △ ACP ≌ △ BCD is known from two equilateral triangles, then BD = AP and Pb = PA + PCPb = BD + DP, so point D is on the straight line Pb

In the acute angle △ ABC, if a point P satisfies PA = Pb = PC, then point P is △ ABC ()
A. Center of gravity

∵ PA = Pb, ∵ P is on the vertical bisector of AB, similarly, P is on the vertical bisector of AC and BC. ∵ point P is the intersection of the three vertical bisectors of △ ABC

Given the curves C1: y = e * x and C2: y = - 1 / E * x, if the tangents of C1 and C2 at P1 and P2 are the same, try to find the tangent equation

Using the geometric meaning of the derivative to do the coordinates of P1 and P2, we can get the tangent equation, Y-Y1 = f '(x1) * (x-x1), where P1 = (x1, Y1); f' (x) is the derivative of the function

If the circle C1: (x-4) 2 + (y + 1) 2 = 9 is inscribed with the circle C2: (x2 + (Y-2) 2 = R2 (r > 0), then

R2=8

The location relationship between circle C1: x ^ 2 + y ^ 2-4x-4 = 0 and circle C2: x ^ 2 + y ^ 2 + 6x + 10Y + 16 = 0 is circle C1 and circle C2
I want the specific process!

(x-2)^2+y^2=1
(x+3)^2+(y+5)^2=18
Center of circle = radical (5 ^ 2 + 5 ^ 2) = 5 radical 2
Radius sum = 1 + 3 root sign 2

If cm = 12, DM = 8, find the length of ab

CD=CM+DM=12+8=20
The radius of the circle is 20 / 2 = 10
OM=CM-CO=12-10=2
OA is radius = 10
In the right triangle OMA, OA = 10 and OM = 2 are known
The length of AM can be obtained, ab = 2am

As shown in the figure, AB and AC are the chords of circle O, AB is vertical, CD and be are the diameters of circle O, if AC = 3, de =?

Connect AE
∵ be is the diameter of circle o
∴∠BAE=90°
∵AB⊥CD
∴AE∥CD
‖ arc AC = Arc de
∴AC=DE
∵AC=3
∴DE=3

As shown in the figure, it is known that AB is the diameter of circle O, CD is the chord, AB is perpendicular to CD, e, of is perpendicular to AC, F, be = of
(1) Verification: of is parallel to BC
(2) It is proved that △ AFO is equal to △ CEB
(3) If EB = 5cm, CD = 10 times root 3cm, let OE = x, calculate the value of X and the area of shadow