Mathematics is difficult in senior one.

20. Set point C as the coordinate origin, CB direction as the positive direction of X axis, CA direction as the positive direction of Y axis, and CC 1 direction as the positive direction of Z axis, and establish a spatial rectangular coordinate system.

(1) according to the meaning of the question: a (0, a, 0) c 1 (0, 0, h) c (0, 0, 0) d (b/2, a/2, 0) b 1 (b, 0, 0).

Let the expression of CDB 1 be x+my+nz=0.

①b/2+am/2=0 ②b+nh=0

So m =-b/a n =-b/h.

Therefore, the expression of CDB 1 is x-b/a*y-b/h*z=0, that is, ah*x-bh*y-ab*z=0.

Distance from point A to plane CDB1= | 0-abh-0 |/√ [(ah) 2+(BH) 2+(AB) 2] = abh/√ [(ah) 2+(BH) 2+(AB) 2]

Distance from point C 1 to CDB 1 = | 0-0-abh |/√ [(ah) 2+(BH) 2+(AB) 2] = abh/√ [(ah) 2+(BH) 2+(AB).

So the distances from point A and point C 1 to the plane CDB 1 are equal.

That is, AC1/plane CDB 1.

(2) according to the meaning of the question: A 1 (0, a, h) B (b, 0, 0) A (0, a, 0) C 1 (0, 0, h)

Vector A 1B=(b, -a, -h) vector AC 1=(0, -a, h).

Because A 1B⊥AC 1 vector A 1B* vector AC 1=0.

That is 0+a 2-h 2 = 0.

a^2=h^2

a=h

When the quadrilateral A 1AC 1 is a square, A 1B⊥AC 1.

2 1、

(1) Set the required straight line l:y=kx+3√2.

Because the chord length =8 and the radius of the circle =5, the distance from the center of the circle to the straight line L = √ [5 2-(8/2) 2] = 3.

|0-0+3√2|/√(k^2+ 1)=3

k= 1

So the equation of the straight line L is y=x+3√2 or y=-x+3√2.

(2) Because the inclination angles of straight line AB and straight line AC are complementary, the slopes of the two straight lines are opposite.

Let's assume that point B is to the left of point A, point C is to the right of point A, and the slope of straight line AB is k, so the slope of straight line AC is-k..

The equation of straight line AB is y-4=k(x-3), and the equation of straight line AC is y-4=-k(x-3).

Substitute in the circle o:

①x^2+(kx-3k+4)^2=25

(k^2+ 1)x^2+(8-6k)kx+(3k-4)^2=25

Because x 1=3, x2 = x1+x2-x1= (6k-8) k/(k2+1)-3 = (3k2-8k-3)/(k2+65448.

y2=k(x2-3)+4=k[(3k^2-8k-3)/(k^2+ 1)-3]+4=4-(8k+6)k/(k^2+ 1)=2( 1-2k)(2+k)/(k^2+ 1)

So point B ((3k+ 1) (k-3)/(k 2+ 1), 2 (1-2k) (2+k)/(K2+1))

②x^2+(4-kx+3k)^2=25

(k^2+ 1)x^2-(6k+8)kx+(3k+4)^2=25

Because x 1=3, x2 = x1+x2-x1= (6k+8) k/(k2+1)-3 = (3k2+8k-3)/(k2+65443.

y2=4-k(x2-3)=4-k(8k-6)/(k^2+ 1)=(-4k^2+6k+4)/(k^2+ 1)=2( 1+2k)(2-k)/(k^2+ 1)

So point C ((3k- 1) (k+3)/(k 2+ 1), 2 (1+2k) (2-k)/(k 2+1))