There are no difficult math proof questions!!! Hurry, hurry

Proof: Pa face ABC, BC within face ABC.

pa⊥bc………

In the circle O, ab is the diameter.

ACB = 90°, i.e., BC AC.........

By and , get.

BC Plane PAC

and ae inside the face pac.

BC AE and AE PC, PC and BC determine the surface PBC AE surface PBC

Proven. Thank you.

Proof: Take the midpoint E of AB and connect to CE

m is the midpoint of A1B, Me is the median of ABA1, ME AA1, and ME=(1 2)AA1 AA1 CC1, and AA1=CC1, and N is the midpoint of CC1 ME CN, and ME=CN

The quadrilateral mecn is a parallelogram.

MN CEMN is not included in the face ABCD.

MN surface ABCD

Thank you.

Take the midpoint e3 of n, m, bb1 as the plane x

Because me a1b1 ab,ne bc and the line me,ne intersect at the point e

So the plane x parallel plane ABCD

And because mn belongs to the plane x

So MN planar ABCD

And it can be made quickly using the coordinate method.

Evidence: as ME AB, vertical foot is E, linked to CE

The hexahedral ABCD-A1B1C1D1 is the cube, and M,N are the midpoints of A1B and C1C, respectively.

em cn quadrilateral cemn is a parallelogram ce mn ce ce in the surface abcd, mn is not in the surface abcd mn plane abcd

af = 2EC (E is the midpoint, similarity ratio 1:2) = 6 Calculated using the Pythagorean theorem: dc = 3:1.

Then use DC to calculate BC

In the right-angled triangle ACD, the Pythagorean theorem can be used to calculate the AC, so that there is a waist length and a base length.

If the bottom midline is 3, do the same.

Do it with similar triangles.

Well, first of all, make a diagram.

It is known that finding mn knows that mn=md-nd

So let's find md and nd, right?

Then ask for MD ad BC first

med∽△mcb

md:ed=mb:cb

ed=ef+fd=2/3ad=32/3

db=20mb=20-md

cb=16 all substitutions.

Get md=8 (calculate the specific process yourself).

Then use the same method to find nd, you can do the math yourself, the same method, tell you that the answer is nd=5 so obviously mn=3

ps: For the sake of my hard work, add some points.

The answer should be 3, you can see from the proportional sides of similar triangles. The triangular EMD and the triangular BMC are similar. The triangle FND and the triangle BNC are similar.

Let lim f(x) = c

For any >0, the presence of a > 0 makes x > a |f(x)-c|For any x > a, by the lagrange median theorem, there exists b (x,x+2), so that f'(b) = (f(x+2)-f(x))/2.

then |f'(b)| f(x+2)-c|+|f(x)-c|)/2 <

In the same way, there is c (x+6,x+8), such that |f'(c)|

Still by the lagrange median value theorem, there exists d (b,c), so that f''(d) = (f'(c)-f'(b))/(c-b).

by c-b > 4, there is |f''(d)| f'(c)-f'(b)|/4 ≤ f'(c)|+f'(b)|)/4 < /2.

by d (b,c) x,x+8), there is 0 < x-d| <8.

By the lagrange median theorem, the existence of e (x,d) makes f'''(e) = (f''(d)-f''(x))/(d-x).

So|f''(x)| = |f''(d)+(x-d)f'''(e)| f''(d)|+x-d|·|f''(e)|2+8( 16) = e > a, hence |f'''(e)| /16).

That is, lim f''(x) = 0.

Similarly, there is g (x,b), such that f''(g) = (f'(b)-f'(x))/(b-x).

By |f''(g)| 0 < x-b|<2, Yes|f'(x)| f'(b)|+x-b|·|f''(g)| 2ε = 3ε.

Also get lim f'(x) = 0.

Rub ah, it's all back to the teacher... I know a 0.,Forget everything else.。。。 The equal sign is also recognized.

Obviously, this is still a testament ... It is directly derived from the differentiable must, and it tends to be constant when there is no lead, and all the guides are 0 (only to 3 me).

The title is wrong. ab is the side of a square. e on bc. So AB is significantly smaller than BC. So AE and EC cannot be equal.

Did you make the wrong drawing or the wrong question? It is clear that AE is not equal to EC square ABCD can know AB=BC, and in the figure AE>AB and EC

Dude, don't think about it, you got it wrong, you can't wait.

ae〉ab ec< BC How could you wait.