Junior 2 Mathematics Parallelogram Please answer in detail, thank you 6 11 48 40

1 3 3 3, the quadrilateral aecd is a parallelogram proof: because ab parallel cd

ab=2cd

E is the midpoint of AB.

So ae = and parallel to cd

So the quadrilateral aecd is a parallelogram.

2. The triangle BCE is equal to the triangle FBA

Proof: Nexus ed

Because ab parallel cd

ab=2cd

E is the midpoint of AB.

So EBCD is rectangular.

So the angle AED is a right angle.

Angle a = 60 degrees.

So ae=af=df=1 2ad

E is the midpoint of AB and E is the midpoint of AB.

So be=af, ab=ad

AECD is a parallelogram.

So the angle A = Angle CEB, CF = Ad = Ab

So the triangle BCE is all equal to the triangle FBA

3. Solution: The triangle BCE is equal to the triangle FBA

So FBA is a right triangle.

E is the midpoint of AB, and the angle A = 60 degrees.

So AEF is a regular triangle.

So the angle AEF = 60 degrees, AF = EF = BE

AB parallel CD, AECD is an equal quadrilateral.

So angle a = angle bec = angle ecd = 60 degrees, angle ecd + angle aec = 120 degrees so angle bef = angle dec

ef=be，cf=cf

Therefore, the triangle ceb is all equal to cef, and it is a right triangle cd=2, so be=2, bc=2, and the root number 3

BCEF can be seen as two right triangles.

Area = 1 2bc * be*2 = 1 2 * 2 * 2 root number 3 * 2 = 4 times root number 3

(1) Parallelogram.

2) BEF FDC or (AFB EBC EFC) Proof: Nexus DE

ab 2cd, e is ab midpoint.

dc eb and dc eb

The quadrilateral bcde is a parallelogram.

AB BC quadrilateral BCDE is rectangular.

aed＝90°

In RT ABE, a 60°, F is the midpoint of AD AE AD AF FD

AEF is an equilateral triangle.

bef＝180°－60°＝120°

and fdc 120°

BEF FDC (SAS).

3) If cd 2, then ad 4, de bc 2 3 s ecf 1 2 saecd cd·de=1 2 2 2 3 2 3

1.Parallelograms (opposite sides are parallel and equal).

2.Triangle CDF triangle BEF congruence.

Auxiliary line: DE, in RTDEA you can use SAS for a lot of implicit conditions

3.Using the area of the trapezoid 8 3 to subtract the area of the two triangles, the base height can be calculated by using the relationship between the corners of 30°60°, and the final result is 6 3

It's pretty easy to do, but it's a bit complicated to describe here.

AECD is simply a parallelogram, and 2) first proves AEDed EBC

And then prove the ABF AED

The result is ABF EBC

3) EFC is actually equal to EBC, so the area is 2*2 (bottom)*2*root number 3 (high) 2

(1) The quadrilateral AECD is a parallelogram. [Both sides are parallel and equal].

2) AED is equal to EDC. [Proof: ae=dc, de is the common edge, aed= edc=90°.] Corner edge].

3) The area of the quadrilateral BCFE is 4 times the root number 3. [The total area of the trapezoidal is (4 + 2) * 2 times the root number 3 2 = 6 times the root number 3. The height is obtained by a sine of 30°.

Subtract the area of the two triangles AEF and FDC, these two triangles are equal in height at the bottom, and the area is 2 * root number 3 1 = root number 3, so in total you have to subtract 2 times the root number 3, so the final area is 4 times the root number 3].

1.The quadrilateral abcd is a parallelogram, ad bc, ad=bc, e, f are the midpoint of ad,bc, de=1 2ad,bf=1 2bc, de=bf, and because de bf quadrilateral bedf is a parallelogram, be df in the same way that the quadrilateral aecf is a parallelogram, af ce quadrilateral ehfg is a parallelogram.

2.Didn't think about it.

(1) Proof:

eab bad bad dac 60°, eab dac, ea da, ba ca, δaeb δadc.

So EBC EBA ABC DCA ABC 120°.

Then EBC BCG 120° 60° 180°, then EB GC, eg BC, BCGE are parallelograms.

2) BEGC is still parallelogram.

Similar to (1), it is easy to prove: δabe δacd, then abe acd 120°, then cbe acb 60°, and then be gc, and bc eg, thus provable.

3) To make it a diamond, you only need to be bc and be cd, so you only need to select the d point to make bc cd.

You're too lazy to give a single figure, but you can still do it:

The first question is a good proof that ab=ac ae=ad (equilateral triangle with equal sides) angle bae is equal to 60 degrees minus angle eac is equal to angle cad

So triangles are congruent (corner edges).

Question 2: A quadrilateral is a parallelogram, and a set of parallel opposites is known, and the degrees of each angle can be calculated.

The third question is established.

The fourth question is that the breakthrough point is that CF is equal to CD

Very rough answer, I hope it helps.

Solution: cd:cf=2:1

cf=cd/2

If the point e is em cf to m, then, em=10 2 cf=20 cf em=20 (cd 2)=40 cd

l1∥l2，ad∥bc

The area of the quadrilateral ABCD = CD em = CD 40 cd = 40

Solution: Let cf=x, ce=y, then dc=2x

ad∥bc，l1∥l2

The quadrilateral ABCD is a parallelogram.

s quadrilateral abcd=cd ce=2xy

s△cef=1/2×cf×ce=10

1/2xy=10

xy=20s quadrilateral, abcd=40

Because the upper and lower lines are parallel. AD BC yields ABCD is a parallelogram, assuming that the height of the parallelogram and CEF is h (the height of the parallelogram and the triangle are the same), s CEF = 10 = CF*h

S parallelogram ABCD = cd*h = 2cf*h = 20

The area of the triangle CEF = 1 2 * CF * CE, and the area of the parallelogram ABCD = CD * CE = 2 * CF * CE = 40

Solution: cd:cf=2:1

s△cef=10

ad bc (the distance between parallel lines is equal, so the high of cde is equal to the high of cef).

s△cde=20

l1//l2 ， ad//bc

The quadrilateral ABCD is a parallelogram.

s parallelogram = 2s cde=40

Answer: The area of a parallelogram is 40

1) Proof of: EAB Bad Bad DAC 60°, EAB DAC, EA DA, BA Ca, ΔAEB δADC.

Yu Tsai is EBC EBA ABC DCA ABC 120°.

Then EBC BCG 120° 60° 180°, then EB GC, eg BC, BCGE are parallelograms.

2) BEGC is still parallelogram.

Similar to (1), it is easy to prove: δabe δacd, simple luck.

Then abe acd 120°, then cbe acb 60°, then be gc, and bc eg, thus prove.

3) To make it a diamond, you only need to be bc and be cd, so you only need to select the d point to make bc cd.

(1) Proof:

ACD and CBF

Because of equilateral ABC

So cb=ac angle, cbf=angle, acd=60 degrees, and because cd=bf

Eliminate the congruence theorem of triangles: if the two sides and their angles are equal, then the triangle is congruent to ACD CBF

2) It seems that the title is not very complete, if there is a picture, it will be clear Landlord, forgive me, I drew a few pictures myself, not the exact answer

1.Equilateral triangle - >ac=cb, b= c, and cd=bf, according to the "corner edge" to get acd cbf

2.That, what is E, and G, can you explain clearly?

How is it the same as my homework?

Where's the dot e? Where's the dot g?

The quadrilateral ABCD is a parallelogram, OA=OC, ob=OD, and af=CE, BH=DG, af-oa=CE-OC, BH-ob=DG-OD, of=oe, OG=OH, and the quadrilateral egfh is a parallelogram, GF HE

Hope it helps

Good luck with your learning and progress o )

In the parallelogram ABCD, bo=od, ao=oc, af=ce, bh=dg

ao+of=oc+eo

bo+oh=od+go

of=eogo=oh

The quadrilateral EGFH is a parallelogram.

gf∥he