Quadrilateral problems in the second year of junior high school, math problems in the second year of

Hello. This question can be solved like this:

Solution: ab parallel cd

fdc=∠fab

AD parallel BC

fab=∠cbe

fdc=∠cbe

ce⊥ab∠ceb=90°

cf⊥ad∠cfd=90°

CFD is similar to CEB

ecb=∠fcd

AB parallel CD

dce=∠ceb=90°

i.e. 1+ ecb=90°

ecf=125°, and ecf= fcd+ ecb+ 1= ecf, i.e., ecf= fcd+ ecb+ 1=125°, then there is 1+ ecb=90° fcd+ ecb+ 1=125°, and substitute fcd+90°=125° to get fcd+90°=125°

fcd=125°-90°=25°

AD parallel BC

cfa= bcf=90°, and bcf= 1+ fcd, this method may be a bit troublesome for me to write.

I don't know if you learn similar triangles, but if you do, you can do this problem. If you haven't learned, you can add some auxiliary lines to do it.

55 degrees. In a quadrilateral EBFC, the sum of the internal angles is 360 degrees, of which two are 90 degrees.

Thus angle 1 and angle b complement each other.

So the angle b is 55 degrees.

Angle 1 and angle b are equal (parallelograms are equal diagonal).

Method 1: Because the folded quadrilateral ABCD is a rectangular piece of paper, AB=AE=CD=6, BF=EF

So you can find ad=bc

Because BF + FC = BC, (the square of BF) - (the square of FC) = (the square of CE), so (the square of AF) = (the square of AB) + (the square of FC) takes the midpoint of AF as the point G

Because BG=eg

So bg=eg=bf=gf (equilateral triangle).

af=2bf

Since ab=2de, the angle EAD is 30 degrees.

The angle AED is 60 degrees.

The angle AEF is 90 degrees.

The angle AFB is 60 degrees.

So the square of (2bf) = the square of af.

4 (square of bf) - (square of bf) = 36

So af = 4 times the root number 3

ae is the folded position of ab--- ae=ab---abe is a regular triangle af bisected bae=60°--baf=30°ab=cd=6---af=6*2 3=4 3

Is your f-spot in**?

Because it is held, ab cd ab=cd because the paper abcd is folded so that point b falls exactly at the midpoint e e of the cd edge so: ae=ab

There will be an answer later. I don't join in the fun!

Solution: Because AEF is ABF folded along a straight line AF, then DE=EF, AE=AB, because CD=6, E is the midpoint of CD, so ED=3, and because AE=AB=CD=6, so EAD=30°, then FAE= (90°-30°)=30°, let FE=X, then AF=2X, in ADE, according to the Pythagorean theorem, (2X)2=62+X2, X2=12, X1=2, X2=-2 (rounded) AF=2 2=4

e is the midpoint, so de=ce=3

Because it is folded, ab=ae=cd=6

In the triangle ade, we can find ad=3 3

Let bf be x, then cf=3 3-x, and bf=ce in the triangle cef, we can find x=2 3, so we can naturally find the value of af in the triangle abf =4 3

Requirements: Make your own drawings.

1. In the known parallelogram abco, one side is 5 and one diagonal is 8, and the value range of the other diagonal is found.

Answer: 2 diagonal 18.

2. In the isosceles trapezoidal ABCD: the waist length is 2, the diagonal is perpendicular to the waist, the upper bottom is equal to the waist, the diagonal is 3 and the number 3 is long, and the degree of the upper and lower angles and the area of this trapezoidal shape are found.

Answer: Top and bottom angle: 120°; Area: 3 roots No. 3

3. In the diamond-shaped ABCD, the point A is the coordinate origin, and the coordinates of the point B are (1, root number 3), and the coordinates of the intersection of the diagonal O are found.

Answer: Point o coordinates: (3 2, root number 3 2).

4. In the rectangular ABCD, AB=CD=9, BC=AD=3, the known points M and N start to move at the same time at a speed of 1M S from A and B respectively, and stop moving when M and N reach the points A and B, and find how much time MDN B is diamond-shaped after the motion.

Answer: After 5 seconds.

Barely make up for a while, add me as a friend, and I'll give it to you next time I'm free, I can't think of it now!

Let the number of edges be n (n is greater than or equal to 3, and n is an integer).

Regular polygons with 2n (n-2) as integers can be tiled.

For example, regular triangles, regular quadrilaterals, regular hexagons, etc. can be tiled, because as long as a point can spell out 360 degrees, it can be tiled.

Triangle... Positive five sides + positive ten sides?

I don't remember the reason ... It seems that what is said and what adds up = 360°.

The proband, EN, is parallel and equal to MG.

It can be converted into a parallel mg of the certificate

It can be converted into bn=dm

From the question easy to prove that the triangle bnf and cgf are congruent, so bn=cd is the same as dm=ae and ae=cd

So bn=dm

That is, proof (I knew you were wrong when I did it, and I can solve it).

I can tell you very clearly, the question is printed wrong, the condition says that ab gf extension line intersects at point n, but it should actually be ab df intersects at point n, and the em=ng that is verified is also incorrect, it should be em=nd, you see if you will not continue to ask me now.

Since f is the midpoint of cb, then cf=fb and gc ab then dcf= nbf and cfd= bfn

So cfd bfn

The same goes for ghm ahe

So there is gm=ae=dc=bn i.e. md=en and md en, so the quadrilateral mend is a parallelogram.

So em=nd

Proving that mg and eb are parallel and equal gives that the quadrilateral mgne is a parallelogram dg=eb

That is, BM=MD is sufficient.

It is sufficient to prove that the triangle MDH is equal to the triangle BMF.

DH = BF again, angle MDH = angle MBF

It is proved by the original proposition that the triangle CGF congruent triangle aeh gets the angle AHE=CFG.

Because of the folding.

So bae= b'ae

Because of the median line fg

So CG=GD

Because ec parallel fg parallel ad

So ef:af=cg:gd

So ef=af

In the RT triangle B'AE.

ef=af=bf

So b'af=∠fb'a

And because of the fg parallel AD

b'ad=∠fb'a

So bae= b'ae=∠b'ad

bae+∠b'ae+∠b'ad=90 degrees.

b'AE = 30 degrees.

Because ab = root number 3

So ae=2

Proof of Burning Dough: Folded.

cde≌△c'de

cd=c'd,ce=c'e，∠cde=∠c'dead‖bc∠c'de=∠ced

ced=∠cdecdce

cdce=c'd=c'e

bc=cd+ad

bc-cd=ad

bc-ce=ad, i.e., be=ad

The quadrilateral difference nano shape is a parallelogram.

abec'It's just trapezoidal, I don't see isosceles)