If line AB is on the same side of plane α and the distances from a and B to α are 3 and 5 respectively, then the distance from the midpoint of AB to α is 0______ ．

If line AB is on the same side of plane α and the distances from a and B to α are 3 and 5 respectively, then the distance from the midpoint of AB to α is 0______ ．

Let AC ⊥ plane α, BD ⊥ plane α, then acdb ⊥ plane α, and PE ⊥ CD through P, then PE represents the distance from P to α. From the plane geometry knowledge, we can see that PE is the median line of trapezoid, so PE = 3 + 52 = 4, so the answer is: 4

If the distances between the two ends a and B of line AB and plane α are 3 and 1 respectively, then the distance between the midpoint C of line AB and plane α is?

2 or 1
When a and B are on the same side of the plane, the distance from C to the plane is half of the sum of the distances from a and B to the plane, (3 + 1) / 2 = 2
When a and B are on both sides of the plane, the distance from C to the plane is half of the difference between a and B, (3-1) / 2 = 1

How to zoom a 1:800 CAD drawing into a 1:1000 CAD drawing?

Enter "SC" to select the enlarged object and press enter. Then select the base point and enter: 1000 / 800 in the scale value. At this time, the figure has become 1:000

How to realize the conversion between three-dimensional graphics in CAD plane graphics

Is it to call up the left view, top view, these views?
View - 3D view - top view (left view) (axonometric view) as you choose

As shown in the figure, a, B, C, D are four points on the circle O. (1) if arc AB = 2 arc CD, try to judge the quantitative relationship between AB and CD, and explain the reason
(2) If AOB = 2, cod

(1) CD < AB < 2CD; the reasons are as follows:
Let 2 arc CD = arc AB = 2m °≤ 180 °,
Take the midpoint e of arc AB and connect EA, EB,
Ψ arc EA = arc EB = arc CD = m °≤ 90 °
∴EA=EB=CD,
In △ EAB, EAB = EBA = (?) 189; m °≤ 45 °,
∠E＝180°－﹙∠EAB＋∠EBA﹚＝﹙180－m)°≧90°＞∠EAB
That is EB ＜ ab ＜ EA + EB
CD＜AB＜2CD
(2) the conclusion and process are basically the same. (1) because angle AOB = 2, angle cod, arc AB = 2, arc CD

It is known that, as shown in the figure, points a, B, C and D are four points on ⊙ o, and ab = CD

It is proved that: ∵ AB = CD, ∵ ACB = ∵ DBC, in △ ABC and △ DCB, ≌ a = ≌ D ≌ ACB = ≌ dbcbc = CB, ≌ ABC ≌ DCB (AAS)

As shown in the figure, P is a point on the line AB, ad and BC intersect at e, ∠ cpd = ∠ a = ∠ B, BC intersects PD at F, ad intersects PC at g, then the similar triangle in the figure has ()
A. 1 pair B. 2 pairs C. 3 pairs D. 4 pairs

∵∵∵ cpd = ∵ B, ∵ C = ∵ C, ∵ △ PCF ∽ BCP. ∵∵ cpd = ∵ a, ∵ d = ∵ D, ∵ APD ∽ PGD. ∵ cpd = ∵ a = ∵ B, ∵ APG = ∵ B + ∵ C, ∵ BFP = ∵ cpd + ∵ C ∵ APG = ∵ BFP, ∵ APG ∽ BFP

As shown in the figure, in the rectangle ABCD, BC = 2Ab, point P is on BC, and satisfies AB + BP = PD, find the value of Tan ∠ APD

Let AB = x, BP = y, then BC = 2x. For the formula of Pythagorean theorem of right triangle PCD sequence, the square of (x + y) = the square of X + (2x-y);
X / y = 3 / 2;
You get tan

As shown in the figure, ab = BC = CD, ad is the chord of ⊙ o, if ∠ bad = 50 °, then ∠ AED=______ ．

Connect OA, ob, OC, OD, ∫ bad = 50 °, ∫ BOD = 2 ∠ bad = 100 °, ∫ AB = BC = CD, ∫ AOB = ∠ BOC = ∠ cod = 12 ∠ BOD = 50 °, ∫ AOD = ∠ AOB + ∠ BOC + ∠ cod = 150 °, ∫ AED = 12 ∠ AOD = 75 °

If two chords AC and BD intersect at point P in circle O, AB is the diameter, AP * AC + BP * BD = AB ^ 2 is proved

It is proved that we take point E on AB so that PE is perpendicular to ab,
Then we can know that the triangle AEP is similar to the triangle ACB, so there is a triangle AEP
AP/PE = AB/AC＝＝＝》 AP*AC = AB*PE (1)
Similarly, the triangle BEP is similar to the triangle bed
BE/BP=BD/AB＝＝＝》 BP*BD=AB*BE (2)
It can be seen from 1 and 2
AP*AC+BP*BD=AB^2