As shown in the figure, in △ ABC, D is the point on AB, and ad = AC, AE ⊥ CD, the perpendicular foot is e, and F is the midpoint of CB

As shown in the figure, in △ ABC, D is the point on AB, and ad = AC, AE ⊥ CD, the perpendicular foot is e, and F is the midpoint of CB

Prove: in △ ACD, because ad = AC and AE ⊥ CD, according to the intersection point of the vertical line and the bottom of the isosceles triangle, we can prove that e is the midpoint of CD, and because f is the midpoint of CB, so EF ∥ BD, and EF is the median line of △ BCD, so EF = 12bd, that is BD = 2ef

X + A / 5-x + A / 10 = 1 {where x is unknown} velocity

No solution
But we can calculate a = 10 / 3

Given a (3,7), B (5,2), the vector AB is translated according to a = (1,2)______ ．

Since AB = (5,2) - (3,7) = (2, - 5), no matter how the vector AB is translated, its length and direction remain unchanged, so its coordinates remain unchanged, so the resulting vector is still ab. so the answer is: (2, - 5)

As shown in the figure, in △ ABC, ∠ a = 60 °, BD ⊥ AC, the perpendicular foot is the point D, CE ⊥ AB, and the perpendicular foot is the point E

It is proved that: (1) BD ⊥ AC, CE ⊥ AB, ∠ AEC = ∠ ADB = 90 ° and ⊥ EAC = ∠ DAB, ⊥ AEC ⊥ ADB, ⊥ aead = acab, ⊥ aeac = ADAB, and ⊥ ead = ∠ cab, ⊥ ade ⊥ ABC; (2) in RT △ AEC, ⊥ a = 60 °, ACE = 30 °, AC = 2ae, ⊥ ade ⊥ ABC, ⊥ aeac = DEBC, that is, ae2ae = DEBC ⊥ BC = 2DE

The distance from a point P on the ellipse x ^ 2 / 25 + y ^ 2 / 9 = 1 to the right focus?
Detailed process!
Online, etc. Online to points, urgent ah!!

If the distance between a point on ellipse (x ^ 2 / 25) + (y ^ 2 / 9) = 1 and the left collimator is 5 / 2, then the distance between P and the right focus?
a=5,b=3
So C = 4, e = C / a = 4 / 5
So the ratio of the distance to the left focus and the distance to the left collimator is 4 / 5
So the distance to the left focus = 2
The sum of the distances to the two focal points = 2A = 10
So the distance to the right focus = 8

As shown in the figure, in the RT triangle ABC, the middle angle ACB is 90 degrees, CD is the height of AB, ab = 5cm, BC = 4cm, AC = 3cm

ab=5,ac=3.bc=4
S=1/2ac*bc=1/2ab*cd
cd=ac*bc/ab=3x4/5=2.4
A: CD length 2.4

Given that the function f (x) = − x + 3 − 3a, x < 0ax, X ≥ 0 (a > 0 and a ≠ 1) satisfies that for any x1, X2 ∈ R, (x1-x2) [f (x1) - f (x2)] < 0, then the value range of a is______ ．

∵ for any x1, X2 ∈ R, (x1-x2) [f (x1) - f (x2)] < 0, the function f (x) is a decreasing function in its domain of definition. Then from the function f (x) = − x + 3 − 3a, x < 0ax, X ≥ 0 (a > 0 and a ≠ 1), we can get 0 < a < 1, and - 0 + 3-3a ≥ A0, the solution is 0 < a ≤ 23, so the answer is (0, 23)]

If the hypotenuse length of a right triangle is 16cm and the inscribed circle radius is 1cm, the perimeter of the triangle is

Make an inscribed circle, draw the radius from the center of the circle to each tangent point, and you can get two right angles. If a + 1 and B + 1 are set respectively, then a + B = 16 (congruent triangle), so the circumference is 34

If a = {a, B}, B = {X / X is a subset of a}, M = {a}, let B be a complete set, then the complement of M is?

A={a,b}
B = {x | x is a subset of a} = {ψ, {a}, {B}, {a, B}}
M={A}={{a,b}}
So the complement of M is {ψ, {a}, {B}
Note: ψ is an empty set
If you don't understand, please hi me, I wish you a happy study!

In the triangle ABC, angle a = 96 ° extends BC to D, angle ABC intersects angle ACD bisector at A2, angle a1bc intersects angle a1cd bisector at A3, and so on
Finding the degree of angle A5

∠BA1C+∠A1BC=∠A1CD 2∠A1CD=∠ACD=∠BAC+∠ABC
SO 2 (∠ ba1c + ∠ a1bc) = ∠ BAC + ∠ ABC
2∠BA1C+2∠A1BC=∠BAC+∠ABC
And 2 ∠ a1bc = ∠ ABC
SO 2 ∠ ba1c = ∠ BAC
Similarly, we can get 2 ∠ ba2c = ∠ ba1c, 2 ∠ ba3c = ∠ ba2c, 2 ∠ ba4c = ∠ ba3c,
2∠BA5C=∠BA4C
So ∠ ba5c = (1 / 2) ∠ ba4c = (1 / 4) ∠ ba3c = (1 / 8) ∠ ba2c = (1 / 16) ∠ ba1c = (1 / 32) ∠ BAC = 96 ° / 32 = 3 °