Who can prove it, help! Planar Geometry .

It is not easy to prove that the size of the arc length can be compared by comparing the height of the arc (n) by comparing the height of the arc

Let the center of the circle be o, the radius be r, and the angle of the center of the circle corresponding to the inferior arc ab is

The variation of the height of inferior arc AB with radius r is discussed.

Inferior arc height = radius of the circle - the height of the OAB on the AB edge.

In OAB there is cos = (2r 2-ab 2) 2r 2

then sin = ab (4r 2-ab 2) 2r 2

s△oab=(1/2)r*r*sinθ=ab√(4r^2-ab^2)/4

Height on ab side = s oab ab= (4r 2-ab 2) 2

Inferior arc height = r - (4r 2-ab 2) 2 = r - [r 2 - (1 4)ab 2].

Discussion: (j).

First of all, for any integers a1, a2, b(a2>a1>b), there are (a2) 2-(a2-b) 2>(a1) 2-(a1-b) 2,a2- (a2-b)< a1- (a1-b), which is not proved here.

Inferior arc height = r 2 - [r 2 - (1 4)ab 2].

When r becomes larger, the above equation r 2 increases, and [r 2-(1 4)ab 2 ] also increases, but the increase of r 2 is smaller than the increase of [r 2-(1 4)ab 2 ], so the value of the global equation becomes smaller;

In the same way, when r becomes smaller, the decrease in r 2 is less than that of [r 2-(1 4)ab 2 ], so the value of the global equation becomes larger.

Therefore, the larger the radius r, the smaller the height of the inferior arc ab corresponding to the fixed chord ab, and the shorter the inferior arc ab corresponding to the fixed chord ab. When the high of the inferior arc AB approaches 0, the inferior arc AB approaches the chord AB. (z)

So when two circles intersect, the length of the inferior arc ab of the big circle is smaller than the length of the inferior arc ab of the small circle.

First of all, according to the formula of arc length l = r2 (l is the arc length, is the radian of the central angle of the arc, r is the radius of the sphere) and the fact that the sum of any two face angles of the three corners is greater than the third surface angle, it is proved as follows:

If the spherical curve a1a2....An is a spherical polyline connected by a large circular arc, and according to the sum of any two surface angles of the three hedral angles is greater than the third angle, the arc A1A2 + arc A2A3> arc A1A3, arc A1A3 + arc A3A4 > arc A1A4 ,... are obtained, arc a1an-1 + arc an-1an > arc a1an, so that the spherical polyline a1a2 is obtained....The length of an is greater than the length of the arc a1an, and the length of the arc a1an is greater than the great circular inferior arc connecting the arc a1 and an (if the two points are the diameter of the sphere, they are equal), the proposition holds.

If the spherical curve ab is not a spherical polyline, take n-1 points on the curve, so that the arc length of each adjacent two points is not greater than the length of the curve 1 n, according to the above conclusion, the length of the polyline obtained is larger than the length of the arc ab, when the number of equinox n is increasing, the length of the polyline will continue to approach the length of the spherical curve ab, and its limit is the length of the spherical curve ab, but the length of the curve must be greater than the length of the arc ab, so the proposition is true.

Plus, there's a lot of reference material in it.

Let the central angle of the chord be 2a, it can be calculated that under the condition of a certain chord length, the inferior arc length is proportional to a sina, and a sina monotonically increases, so when the central angle is larger, the arc is longer.

I've seen it. But I don't get that thing in n years.

It's very simple, anyone who has studied mathematics knows it, you pig

I don't know if it's really not a mother's raising

I'm telling you, I don't know anymore haha!

It's garbage.

If this problem only proves that the surface EFG and the vertical plane PDC are redundant, then the condition of AD=PD=2MA is redundant.

Proof: MA is perpendicular to the square ABCD plane, while PD parallel MA PD is perpendicular to the square ABCD plane.

then PD DC, PD BC

The PDC plane is perpendicular to the square ABCD plane.

BC DC again

The BC is perpendicular to the PDC plane.

and g and f are the midpoints of pb and pc, respectively.

GF BCGF is perpendicular to the PDC plane.

EFG vertical plane PDC

I don't think I need to mark the proof theorem).

Untie; Because pd=ab abcd is a square.

So pd=ab=bc=cd=da

Connect the PC PC inside the planar PBC.

Because pd=dc

Therefore, the angle between PC and DC is 45 degrees.

Let the point A1 make A1A=PD=DC=CB=BA=AD and A1A perpendicular ABCD

Connect A1A, A1B, A1P

Because A1A=A1B=PD=DC and both PD and A1A are perpendicular to ABCD, PC=A1B

Connect to the DB AC

Because ABCD is a square.

So db=ac

Because db=ac, pd=a1a, and pd and a1a are both perpendicular to abcd, and because dc=cb=ac

So db=ac

And because db=ac pd=a1a

So bp=a1c

And because PC=A1B, Pa1=BC, PB=A1C, PA1BC is rectangular.

Because PA1BC is a rectangle and the point E is the midpoint of PB.

So PE=BE=A1E=EC

Point E is the center point of the plane A1BCP.

Set point E1 on PC and point E2 on A1B

Connect to e1e2

Make E1E2 cross point E and be perpendicular to PC and A1B

Because point E is the midpoint between A1C and PB.

And the vertical E1E2 passes through the point E, and the vertical is connected to PC and A1B, so E1E=E2E=one-half BC=one-half Da is connected to E1D

Because PD=DC and the angle CPD=Angle PCD

So e1c = e1p

E1P is perpendicular to planar PA1BC

Set the point x on the AD line to connect EX

Make the XE perpendicular plane PA1BC

ex=de1 e1e=dx

Because ex=de1, e1e=dx and dx=ax and the point x is on the ad line.

XE vertical plane PA1BC

Because PA1BC is the same plane as PBC.

So XE is perpendicular to the planar PBC

and x is in the pad plane.

i.e. point x is the point f sought

Therefore, EF is perpendicular to PBC, and EF belongs to the surface EFG, so EFG is perpendicular to the surface PBC, so the surface EFG is perpendicular to PDC

Napoleon's theorem corollary 2

Make triangles BCP, CAQ and ABR on the outside of ABC so that PBC= QAC= , PCB= QCA= , Rab= RBA= , and + 90°, then RQ=RP, and QRP=2

Prove. rb rotates 2 counterclockwise around r to rg, with bg, ag, qg gba = gbr-

Ra=RB=RG, i.e., R is the outer heart of ABG, ABG ACQ BCP, Bac= GAQ, and RGQ= AGQ+ AGR

abc+α+rbp，∴∠rgq≌△rbp．∴rq=rp．

And because grq= brp, qrp= grb=2, this problem is about to substitute =45°, 30°, 15°.

Look at the picture to speak,,Circle problem.

Proof: Because of AD

becf is the three middle lines of the triangle abc, so that ad, cf, be intersect o, fg intersect ad, then o is the center of gravity of the triangle, from the nature of the center of gravity, it can be obtained: oe=1 2ob, of=1 2oc because af=bf, fg be, ab eg so fk=1 2ob, fgeb is a parallelogram so fk=oe, fg=be

So ob=kg

Since of oc=1 2, fk kg=fk ob=1 2, the inference from the parallel lines obs: ok gc

So ad gc