Ask two questions about plane solid geometry~ 1. In cube AC1, G and H are the midpoint of BC and CD respectively, and prove that D1, B1, G, h and four points are coplanar 2. Cube ABCD -- a1b1c1d1, e is the midpoint of AB, proving that CE, d1f, DA are three lines in common Please help me, great Xia,

Ask two questions about plane solid geometry~ 1. In cube AC1, G and H are the midpoint of BC and CD respectively, and prove that D1, B1, G, h and four points are coplanar 2. Cube ABCD -- a1b1c1d1, e is the midpoint of AB, proving that CE, d1f, DA are three lines in common Please help me, great Xia,

1. It is proved that: GH is the median line on the BD side of △ DBC, so: GH ‖ DB is in the quadrilateral dd1b1b, because: dd1 ‖ BB1, dd1 = BB1, so: the quadrilateral dd1 B1B is a parallelogram, so: DB ‖ d1b1, so: GH ‖ d1b1, so; the straight line GH and d1b1 determine a plane

Mathematical solid geometry problems
In tetrahedral s-abc, ∠ BAC = 90 ° and ∠ SAB = ∠ sac = 60 °
(1) When SA = a, find the projective length of SA in plane ABC,
(2) Find the angle between SA and plane ABC

(1) a; (2) 45°
The results are as follows: (1) in the plane ABC, make the so ⊥ plane ABC at O, in the plane ABC, make the OE ⊥ AC, of ⊥ AB, the ⊥ SAB ≌ sac = 60 °, the ≌ ASE ≌ ASF, the ≌ o point on the bisector of the ≌ bac, AE = sacos60 ° = a, Ao = = a;
(2) In RT △ Sao, cos ∠ Sao = =,  Sao = 45 degree

Mathematical solid geometry of the same plane
The parallelogram ABCD is a plane. If you rotate the whole figure up and down a little, it becomes a parallelogram a.b.c.d. then the parallelogram ABCD and a.b.c.d. are not the same plane, right?
It's not going up

Do you use the visual angle of the eye, or the actual parallelogram
Eyes, if you do it on the drawing, you will not see a plane
If it's actually rotating around the center of the figure, it's a plane

As shown in the figure, AB is the diameter of ⊙ o, ab = 2, OC is the radius of ⊙ o, OC ⊥ AB, point D is on AC, ad = 2CD, point P is a moving point on radius OC, and the minimum value of AP + PD is obtained

As shown in the figure, connect BD and ad. it is known that B is the symmetry point of a with respect to OC, so BD is the minimum value of AP + PD, ∵ ad = 2CD, and the degree of arc AC is 90 ° arc, ∵ ad is 60 ° arc, so ∠ B = 30 °, ∵ AB is diameter, ∵ ADB = 90 ° and ab = 2, ∵ BD = 3. Therefore, the minimum value of AP + PD is 3

As shown in the figure, PA and Pb are tangent to ⊙ o at two points a and B respectively, and op = 2, ∠ APB = 60 °. If point C is on ⊙ o, and AC = 2, then the degree of circumference angle ∠ cab is______ ．

Connecting AB, ∵ PA and Pb are tangent to ⊙ o at two points a and B respectively, and ∵ APB = 60 °, ∵ Pao = ∵ PbO = 90 °, ∵ OPA = 12 ∵ APB = 30 °, ∵ AOB = 360 ° - ∵ Pao - ∵ PbO - ∵ APB = 120 °, ∵ OA = ob, ∵ OAB = ∵ oba = 180 °− aob2 = 30 °, ∵ OP = 2, ∵ OA = 12op = 1; ∵

It is known that PA and Pb are tangent lines of circle O, PCD is secant, and ac * BD = ad * BC is proved
The graph is about a circle. From top to bottom, AP is tangent, PCD is secant, Pb is tangent, and then AC, CB, BD, ad (BD does not pass through the center of the circle)

PA tangent circle O to a
So angle PAC = angle PDA
So triangle PAC is similar to triangle PDA
So AC / ad = PC / PA
Similar triangle PBC similar triangle PDB
So BC / BD = PC / Pb
Because PA and Pb tangent circle O to a and B
So PA = PB
So AC / ad = BC / BD
So ac * BD = ad * BC

It is known that PA and Pb are tangent lines of O, and PCD is secant. It is proved that AC times BD = ad times BC
Sorry, I can't draw. I hope you can try,

Because there is a common angle ∠ APC, ∠ PAC = ∠ PDA, we can get △ PAC ∽ PDA, so PA: PD = AC: ad. similarly, we can get △ PBC ∽ PDB, Pb: PD = BC: BD, and PA and Pb are tangent lines of ⊙ o, then PA = PB, so AC: ad = BC: BD, so AC × BD = ad × BC

As shown in the figure, PA and Pb are the two tangents of circle O, AB is the tangent point, PCD is the secant of circle O. through C, make a straight line, AB intersects e, ad intersects F, and CE = EF
Verification: CE / / PA

Please give me the picture!

As shown in the figure, BC is the diameter of ⊙ o, P is a point on ⊙ o, a is the midpoint of arc BC, ad ⊥ BC, perpendicular foot is D, Pb intersects AD and AC at points E and f respectively, is AE equal to be? Why? It stops after 20 minutes

It's late. It's not equal

As shown in the figure, it is known that BC is the diameter of ⊙ o, ah ⊥ BC, perpendicular foot is D, point a is the midpoint of BF, BF intersects ad at point E, and be · EF = 32, ad = 6. (1) prove: AE = be; (2) find the length of de; (3) find the length of BD

(1) It is proved that AF, AB and AC are connected because a is the middle point of BF, and ∠ Abe = ∠ AFB. AFB = ∠ ACB, and ∠ Abe = ∠ ACB. ∵ BC is the diameter, and ∠ BAC = 90 ° ah ⊥ BC. ∵ BAE = ∠ ACB. ∵ Abe = ∠ BAE. ∵ AE = be. (3 points) (2) let de = x (x > 0), from ad = 6, be · EF = 32, AE · eh = be · EF, (4 points), then (6-x) (6 + x) = 32, we get x = 2, that is, de = x (x > 0) In RT △ BDE, BD = 42 − 22 = 23. (7 points)