Junior high school math, I m in a hurry, hurry up Thank you

1.Proof: Since it is known that the triangle ABC is an equilateral triangle, so the angle A = angle B = angle C = 60 degrees, and because de BC, the angle ADE = angle B = 60 degrees (two straight lines are parallel, the isotopic angle is equal) The same reason is that the angle AED = angle C = 60 degrees, so the angle A = angle ADE = angle AED = 60 degrees, so it can be known that the triangle line ADE is an equilateral triangle.

2,.The triangle is similar, d is the midpoint, and the triangle ade is similar to the triangle abc, ab=, ad=, de=, bc=2de=

1。Because de is parallel to bc, the angle d is equal to the angle b, so the angle d is equal to the angle b is equal to sixty degrees, and all sixty degrees are equilateral triangles.

ac=2x, x square plus 4x square equals square, de equals half of bc, or you can do it once as before.

Question 1. ABC is an equilateral triangle.

a=∠b=∠c=60°

de//bc

ade = b = 60°, aed = c = 60° (isotopic angle theorem) a= ade= aed = 60°

ADE is an equilateral triangle.

Question 2. bc⊥ac

sina=bc/ab

again a=30°

sina=1/2

bc/ab=1/2

AB=BC=BC AC, DE AC

ade≌△abc

de/bc=ad/ab

Point d is the midpoint of AB, i.e., ad AB=1 2DE BC=1 2

de=bc*1/2=

Question (1): Because DE is parallel to BC, the angle ADE = angle B 60 degrees, and the angle AED = angle C 60 degrees, so: ADE is an equilateral triangle.

Question (2): The right-angled side of a right-angled triangle at 30 degrees BC is half the hypotenuse), de median line theorem).

Solution: 1) ABC is an equilateral triangle.

All three corners are 60°

And because de bc

ade=∠aed=60°

a=60°So: ade is an equilateral triangle.

1>. The isotopic angle of the parallel lines is equal.

The short side of the angular rt triangle is half the hypotenuse.

Oh health is important in the mentality, to control the mentality OK, please think about it.

1.Let the endpoint of the rectangular middle line be .

ab=dc=1 ad=bc=ef is known

ab+dc+ad+bc+ef=6

So ad=bc=ef=4 3

So the area of the rectangle s=ab*bc=1*4 3=4 3 2Let ab=m

Knowing ab=dc=m ad=bc=(6-3m) 3=2-m, so the area of the rectangle s=ab*bc=m*(2-m) and 0 Question 3 What is the total length?

Question 4 What about the picture?

It is recommended to add a math teacher***, which will definitely solve your problem.

Just ask your classmates

1. 4/3

2.Your picture may not be complete, is your question the 23rd question of the 2006 Zhejiang Jinhua City High School Examination?

(1) From the question, you can know the abc cde

bc=cd,ce=ac,∠bca=∠ecd∵∠ace=∠eca

bce=∠acd

BCE ACD (Corner Edge).

2）△bcm≌△dcn

abc ced and are equilateral triangles.

BM vertically bisects AC, DN vertically bisects CE

bmc= dnc, cm=cn, bc=dc, so bcm dcn

(1) Because abc and cde are equilateral triangles, bc=ac, ce=cd, bca= ecd=60°, ace= eca=60°

bce=∠acd=120°

BCE ACD (Corner Edge).

2) BCM ACN, ECM DCN Certificate: BCM ACN

ABC and CED are equilateral triangles, BCE ACD BC=AC, BCM= ACN=60°, CBM= CAN BCM ACN

3) Still holds.

(1) Equilateral abc and cde bc=ac ce=cd ecd= acb=60° ecd+ ace= acb+ ace so bce acd (asa) (2)dnc cme bmc anc verify: dnc cme prove: because abc+ ace+ ecd=180° and abc= ecd=60° ace=60° is obtained by (1).

bce acd bec= cda and ce=cd dnc cme (3)The graph cannot be drawn on the computer, the answer is yes. You can just ask for it according to my method, and the solution is the same! Pure hands to give points!

(1) 9x +12xy+4y, where x=4 3, y=-1 2:

3x+2y)^2

x=4/3,y=-1/2:

Original = (3*4 3-2*1 2) 2=9

2)(a+b 2) -a-b 2), where a=-1 8, b=2

a+b/2+a-b/2)(a+b/2-a+b/2)=2ab

a=-1/8,b=2.

Original formula = -2*1 8*2= -1 2

2x²+2x+1/2:

1/2(4x²+4x+1)

1/2(2x+1)^2

x+1)(x+2)+1/4.

x²+3x+2+1/4

x²+3x+9/4

1/4(4x²+12x+9)

14(2x+3)^2

Factor it first, and then calculate the evaluation.

1)9x²+12xy+4y²

Original = (3x+2y).

Substituting x=4 3,y=-1 2 into the original formula.

3x4/3+2x(-1/2)）=3

2)(a+b/2)²-a-b/2)²

Original = a - (b 2).

Substituting a=-1 8 and b=2 into the original formula.

1 8) -2 2) =-65 64 Factoring the following formulas:

1）2x²+2x+1/2

Original = 2 (x + x + 1 4).

2(x+1/2)²

2)(x+1)(x+2)+1/4

Original = x +3x + 2 + 1 4

x²+3x+9/4

x+3/2)²

It's easy. I'm also a sophomore in junior high school. Just learned this hehe!

1)(a-b)/(a+b)*(a^4-a^2b^2)/a^2-ab=(a-b)/(a+b)*(a^2-ab)(a^2+ab)/a^2-ab

a-b)/(a+b)*(a^2+ab)=(a-b)/(a+b)*a(a+b)

a(a-b)=a^2-ab

2) Direct = (2x-y) 2 (2x+y)*1 4x 2-y 2=(2x-y) 2 (2x+y)*1 (2x+y)(2x-y)=2x-y (2x+y) 2

2x-y/4x^2+4xy+y^2

Because you didn't put brackets, I just wrote it as I thought.

There may be something wrong, but you didn't put parentheses, and I can't help it.

It's late. I slept... No need to ask.

Finally, I will tell you about the process of multiplication and division of fractions.

1. Factoring first (if possible).

2. Simplification is the same as fraction multiplication and division. For example, if the numerator and denominator are divided by a at the same time, then it is 1 b.

3. It's the result. It's actually quite simple. Try to do it yourself. I slept... Don't ask...

This question is very easy to do.

As an OE vertical OC

Since ab is the tangent of o.

So the angle ODB = angle oec = 90 degrees.

by ab=ac

So angular dbo = angular eco

and ob=oc

So you can get the congruence (you write it yourself).

So od=oe

and dd on the circle.

So e is also on the circle.

So ac is the tangent of o.

Solution: Because ad bisects bac and bac=60°, bad= cad=30°

Because def=60°

So afe= def- bad=30°

Because ae=ce

So afe= bad= cad= eca=30° and because ae=ae

So AEF AEP

So ac=af=8

Extend CE with AB and point G

It can be concluded that CE is perpendicular to AB

e is the vertex as a 60-degree angle ab is the point f? Which angle is 60 degrees?