In the parallelogram ABCD, E F is the midpoint of AB and CD, respectively.

Proof: Parallelogram ABCD

a=∠c,ad=cb,ad=bc

e and f are the midpoints of the edges ab and cd, respectively.

ae=cf△ade≌△cbf

ade≌△cbf

de=bfPlease praise it.

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Proof: In the parallelogram ABCD, A= C, Ad=BC, E, F are the midpoints of AB and Cd, AE=CF A= C, AE=CF

In AED and CFB, AD=CB, AED CFB(SAS).

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abcd so dfeb, because e, f is the midpoint, dc = ab so df = eb, so sun xiao thought that the parallel quadrangle was dismantled. It can be seen that ADE is an equilateral triangle, so DE=AE=EB=BF=FD, so DFBE is a prism, and the circumference is 4 2=8,2,1· Proof: Because the quadrilateral ABCD is a parallelogram.

So dc is parallel and equal to ab

Because of the regret of the jujube e, f are ab, cd midpoint, respectively.

So df is parallel and equal to be

So the quadrilateral dfbe is a parallelogram.

2, because ad=ae=2 and e is the midpoint of the edge ab, ae=be=2

Verify that the quadrilateral DFBE is a parallelogram.

If AD=AE=2, A=60°, find the perimeter of the quadrilateral EBFD.

In the parallelogram ABCD, AB=CD, AB CDE, and F are the midpoints of AB and CD, respectively.

BE=FC quadrilateral, BCFE is a parallelogram.

ef=bc

Answer: The quadrilateral abcd is a parallelogram, ab dc, ab=dc, and e and f are the midpoints of ab and dc, ae=fc, and ae fc, and the quadrilateral aecf is a parallelogram a set of quadrilaterals that are parallel and equal to the opposite sides are parallelograms, baf= ecd.

So angular edc = angular fce

and e and f are the midpoints.

So fc=ed

EC common edge, triangle FEC, congruent triangle DCE

So ef=cd

So the quadrilateral fedc is a parallelogram.

EF Parallel CD

AB parallel CD again

then there is AB parallel EF

So EF ad BC

Because ae=1 2ab=1 2cd=df, and ab cd, the quadrilateral aefd is a parallelogram, and ad ef is obtained;

In the same way, it can be proved that the quadrilateral EBCF is a parallelogram, and EF BC is obtained

So EF ad BC

Proof: The quadrilateral ABCD is a parallelogram.

ae‖cf，ab=cd

E is the AB midpoint and F is the CD midpoint.

AE=CF quadrilateral, AECF is a parallelogram.

The same can be done for af ce, and de bf can be obtained

The quadrilateral ggfh is a parallelogram.

Yes, it can be proved that the triangle age and the triangle HFC are congruent triangles, and then it can also be proved that HF is parallel to GE, because GE is parallel and equal to HF, so EHFG is a parallelogram.

Proof: The quadrilateral ABCD is a parallelogram.

ae‖cf，ab=cd

E is the AB midpoint and F is the CD midpoint.

AE=CF quadrilateral, AECF is a parallelogram.

The same can be done for af ce, and de bf can be obtained

The quadrilateral fgeh is a parallelogram.

Give me extra points!!

You must be a high school student, he's a parallelogram. The proof is as follows;

AE is parallel and equal to CF, AECF is a parallelogram, and it is concluded that GF is parallel to EH. In the same way, it can be concluded that EG is parallel to FH, which shows that EFGH is a parallelogram.

Personally, I think it's right, just take a look.

(1) Proof: In the parallelogram ABCD, AD=BC, D= B, and DAF= BCE

daf≌△bce（asa）．

2) Solution: The sum of the internal angles of the quadrilateral QCFM is 360°, abc=60°, ecb=20°, bec=100°, daf bce, be=df, ae=cf, ab cd, the quadrilateral aecf is a parallelogram, eaf= bec=100°, aec= mfc=80°, then qmf+ mfc+ fcq+ cqm

amn+80°+100°+50°=360°∴∠amn=130°．

(1) Solution: The quadrilateral BCD is a parallelogram.

AD is parallel and equal to BC

d=∠b∵∠daf=∠bce

daf≌△bce（asa）

2) Solution: abc=60°

bad=120°

daf≌△bce

daf=∠bce=20°

baf=100°

bn split abc equally

abm=30°

amb=50°

amn=130°

Connecting EF, E and F are the midpoints of AB and CD, so AE=DF and AE DF, so AEFD is a parallelogram.

g is the intersection of the diagonals, so g is the midpoint of af. In the same way, h is the midpoint of fb. That is, to do mammoth GH is the median line of the pure clear bridge of the triangle AFB.

GH parallel AB, GH = 1/2 AB

Because E and F are the buried points of AB and CD respectively, EF is parallel to AD and BC and parallel to the Zhiwang quadrilateral ADEF

BCFEG is the diagonal intersection.

So ag=gf

Similarly. bh=hf

So GH is the triangular ABF median.

So gh parallel AB, GH = 1/2 AB