Kneel and beg!! A math proof problem! Ask for the specific process!!

Proof: Because ABCD is rectangular, AB=CD, and because it is folded in half and vertically on the tabletop, both AB and CD are perpendicular to BC, so AB CD, so ABCd is rectangular after verticalization, so AD=BC and AD BC.

2) Because EF is the discount of ABCD after the discount, there is AE=BF DE=CF and AD=BC is obtained from (1), so the triangle ADE is all equal to the triangle BCF All angles ADE=angle BCF

Note) just change the words above into symbols.

Prove that EF is a rectangular polyline.

ae∥bf de∥cf ab∥ef∥dc

ae∩de=e bf∩cf=f

Planar ADE Planar BCF

Planar ABCD Planar ade=AD

Planar ABCD Planar BCF = BC

AD BCACBCD is a parallelogram.

AD BC and AD = BC

de∥cf ad∥bc

ade=∠bcf

Solution: 1) Derived from the question.

CD = AB, CD parallel AB

CD = AB, CD parallel AB

The quadrilateral ABCD is a parallelogram.

ad is parallel and equal to cd

2) Derived from the title.

De parallel CF AE parallel BF CD parallel EF parallel ab can prove that the quadrilaterals CFF and ABFE are parallelograms.

Finally, we can get de=cf, ae=bf

So AED is equal to CFB

So ade= bcf

Fold in half, so the triangle ADE and BCF are congruent, (with SAS certificate) 1,2 is derived.

Read more books and do more hands, it's much faster than asking for answers online.

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.

AE bisects BAC and ED bisects ADC

bad+∠adc=180°

dae=1/2∠bad、∠adc=1/2∠adc∠ead+∠eda=90°

aed=90°

The same can be said for BGC= GFE=90°

The quadrilateral efgh is rectangular (there are three corners that are 90° and the quadrilateral is rectangular).

It's very simple The outer angle of the rectangle is also 90 degrees Ah The angle bisector line, after the intersection of the bisecting angle, the top angle of this triangle is 90 degrees, so this quadrilateral EFGH must be rectangular, and then only the four sides are required to be equal, this is also easy to get The vertical distance from the angle bisector to the two corner sides is equal, that is, the distance from each vertex of the quadrilateral EFGH to the outer corner edge of the rectangular ABCD is equal and the opposite side of the rectangle is equal, you can get EA=EB=GD=GC; fb=fc=ha=hd;You can get that all four sides are equal. It can be proved that the quadrilateral efgh is a square. Okay, think about it yourself, ok, thank you.

To simplify ln(x1 x2)=(x1-x2) (x1+x2)ln(x1 x2)=[(x1 x2)-1] [(x1 x2)+1] so that x1 x2=t

Then lnt=(t-1) (t+1)=1-[2 (t+1)] I forgot the specific graph, the left is the inner logarithmic function, the right is the hyperbolic function, then the two curves only allow one intersection point, and it is easy to know that when t=1 the equation holds, then we can know that t is not equal to 1 when the equation is not true, that is to say, if and only if x1=x2 the equation holds, otherwise the equation does not hold.

x1, x2>0, then t=x1 x2, so that f(t)= ln(t)-2(t-1) (t+1).

f ' t) "Derivative" = 1 t -4 [(t+1) 2] = [(t-1) 2] [t(t+1) 2] t>0

then f'(t) >= 0

then f(t) is a monotonic function at t>0.

t=1 f（t）= 0

t is not equal to 1 then f(t) is not equal to zero.

Replace t with x1 x2 and the problem proves f

Just give a counter-example:

Let x1=e, x2=1

inx1-inx2=1

2(x1-x2) (x1+x2)=2(e-1) (e+1) of course the two are not equal.

Proof: Known AD BC

So cad = acb (two parallel lines with equal inner wrong angles) i.e. ead= fcb

Known de bf

So efb= fed (two lines parallel with equal inner wrong angles) and efb+ bfc=180

fed+∠aed=180

So aed= bfc

AF=CE is known

So af-ef=ce-ef

i.e. ae = cf

ead = fcb (calculated).

So triangle aed

Triangular CFB (ASA).

So ad=cb (the corresponding sides of the congruent triangle are equal).

And AD BC

So the quadrilateral abcd is a parallelogram (a quadrilateral with parallel and equal opposite sides is a parallelogram) so ab=cd (the opposite sides of the parallelogram are equal).

ad bc (known), cad = acb (two lines are parallel, the inner wrong angles are equal) de bf (known), def= bfe (two lines are parallel, the inner wrong angles are equal) af = ce (known), ae = cf (equal minus equal amount, equal difference) aed cfb (corners and corners) ad=bc (congruent triangle corresponding sides are equal) ad bc (verified).

AD BC (known).

The quadrilateral ABCD is a parallelogram.

ab and dc

ad∥bc，de∥bf

dae=∠bcf

dea=∠bfc

af=ceae=cf

According to the corner edges, the triangle DAE and BCF are congruent.

AD = BC (they are also parallel).

So ABCD is a parallelogram.

So AD and BC are parallel and equal.

I can't see your diagram, but it seems to be quite simple, and I tried to draw it myself, and it should be like this

Proof: Connect bp, pass p to make bn's perpendicular line, and the perpendicular foot is nThe perpendicular line of the BM is made over P, and the perpendicular foot is FThe perpendicular line of P to do AC is D

Because Pa and PC are the MAC of ABC and the angular bisector of NCA, respectively, PN=PD and PD=PM

i.e. pn=pm, so bp is the angular bisector of mbn.

The theorem used is that the distance from one point to both sides of the bisector is equal

Junior high school questions, I'm already in college, and it may be a bit tedious to write, but it's good to inspire you.

Answer: eg=12 2=6 (6 2+8 2=10 2) eg 2+dg 2=de 2 , edg is a right triangle dg ef, and g is the midpoint of ef, def is an isosceles triangle (three lines in one) I hope it can be helpful to you, thank you for adopting.

It is not difficult to prove that deg is a right triangle, that is, there is dg perpendicular eg, so it is not difficult to deduce def as isosceles.

Let's start with a lemma: the midline length theorem.

It can also be certified by passing C to do CE parallel DB AB extension line to E.