A is a point on the circle O with the diameter of BC. Passing through point B, the tangent line of circle O intersects with the extension line of Ca at point D, e is the midpoint of BD, and the extension AE intersects with CB at point F If sin ∠ f = 3 / 5, calculate the value of sin ∠ D

A is a point on the circle O with the diameter of BC. Passing through point B, the tangent line of circle O intersects with the extension line of Ca at point D, e is the midpoint of BD, and the extension AE intersects with CB at point F If sin ∠ f = 3 / 5, calculate the value of sin ∠ D

Connect AB, oadb tangent circle O at point B, BC is the diameter ﹥ DB ⊥ FC at B ﹥ FBE = ﹥ DBC = 90 ° and ﹥ BAC is the circumference angle of diameter BC ﹥ BAC = 90 ﹥ DAB = 180 ° - 90 ° = 90 °

As shown in the figure, P is in ∠ AOB, and points m and N are the symmetrical points of point P about AO and Bo respectively, and the intersection points E and F with AO and Bo. If the perimeter of △ PEF is 15, the length of Mn is calculated

The ∵ point m is the symmetric point of point P with respect to Ao, ∵ Ao bisects MP vertically, ∵ EP = em. similarly, PF = FN. ∵ Mn = me + EF + FN, ∵ Mn = EP + EF + PF, the perimeter of ∵ PEF is 15, ∵ Mn = EP + ef + pf = 15

As shown in the figure, in trapezoidal ABCD, ad ∥ BC, m and N are the midpoint of AD and BC respectively, and E and F are the midpoint of BM and cm respectively. (1) prove that quadrilateral menf is parallelogram; (2) if quadrilateral menf is diamond, what conditions need to be added in trapezoidal ABCD? Please write down this condition

(1) It is proved that: in △ MBC, N, e, f are the midpoint of BC, BM, cm respectively, ∥ en ∥ MF, ∵ en = MF, ∥ quadrilateral menf is a parallelogram. (4 points) (2) if the quadrilateral menf is a diamond, the condition AB = CD. (6 points) (the answer is not unique, other answers are given points)

Let a be the moving point on the circle (x-1) ^ 2 + y ^ 2 = 1, PA be the tangent of the circle, and PA = 1, then the trajectory equation of point P is

Drawing ~ you will find that the tangent PA, the radius of circle a and the center of point P to circle a form an isosceles right triangle. The length of right angle side is 1, and the length of hypotenuse side is root sign 2. According to the definition of circle: the set of points with the same distance to a fixed point. Therefore, point P forms a circle with (1,0) as the center and half diameter r = root sign 2. The circle equation: (x-1) ^ 2 + y ^ 2 = 2

If we know m (0, - 1), n (0,2), and the moving point P satisfies PM PN = 3, then the trajectory of P is,

It's a ray, x = 0, Y > 2, because Mn = 3

The following equation is transformed into an algebraic expression of one unknown to represent another
1.4x-y=-1
2.5x-10y+15=0

x=(-1+y)÷4
y=1+4x
x=(-15+10y)÷5
=-15÷5+10y÷5
=-3+2y
y=(15+5x)÷10

Let the volume of the triangular prism abc-a1b1c1 be V, P and Q be the points on the side edges Aa1 and CC1 respectively, and PA = qc1, then the volume of the pyramid b-apqc is (V / 3)
Let the distance from B to plane AC1 be h, and the area of parallelogram a1acc1 be s
The volume of triangular prism abc-a1b1c1 is v = SH / 2
PA = qc1, APQC area = a1acc1 area / 2 = s / 2
The volume of pyramid b-apqc = (s / 2) H / 3 = SH / 6 = V / 3
Why is the volume of triangular prism abc-a1b1c1 v = SH / 2
Divide by 2. Isn't the volume of the prism base set times the height

1 pass B to make the height on the side of AC, and cross AC to d
SPAQC=1/2SAA1C1C
VB-PAQC=1/2SAA1C1C*AC*BD*1/3
VABC-A1B1C1=SAA1C1C*BD*1/2
Therefore, vb-paqc = V / 3
V=SH/2
Because the area is saa1c1c and the height is BD

Level 2 square sequence 9,16,36100 (), why is 324 in brackets
I don't really understand~
I want to know why is 18 square ~ I compare benzene ~ really don't understand ~ sorry~

9 = 3 & sup2; 16 = (3 + 1) & sup2; 36 = (3 + 1 + 2) & sup2; 100 = (3 + 1 + 2 + 4) & sup2; the next number should be (3 + 1 + 2 + 4 + 8) & sup2; = 18 & sup2; = 324, the first number is the square of 3, the second number is the square of 4-3 = 1, the third number is the square of 6, 6-4 = 2, 2 = 1 × 2, the third number is the square of 10-6 = 4

Given the set a = {x, XY, x + y}, B = {0, the square of X, y}, if a = B, find the value of real number x, y

When a = B (1) x = 0, elements a and B repeat (2) the square of x = x & nbsp; when x = 0, elements repeat (2) the square of x = x & nbsp; when x = 1, a = {1, y, y + 1} B = {0,1, y} when y = 0, elements repeat y + 1 = 0, y = - 1, satisfying (3) & nbsp; when x = y & nbsp; xy = the square of X. therefore, when x + y = 0, the solution is x = 0, it is obvious that elements a and B of the set

Given 1 = 2 of a + A, find the value of a & # 178; + (1 / A & # 178;)
Such as the title

A + 1 / a = 2
（a+1/a）²=2²
a²+2+1/a²=4
a²+1/a²=2
If other problems are added,