1, (Taiyuan, Shanxi, 2008) As shown in the figure, in the rectangular ABCD, diagonal AC and BD intersect at point O, and it is known that AB=2.5, then the length of AC is.

2. (Xiaogan, Hubei, 2008) Four congruent right-angled triangles form a big square, and the vacant part in the middle is a small square, thus forming a "Zhao Shuangxian diagram" (as shown in the figure). If the area of the small square is 1, the area of the large square is 25, and the smaller acute angle in the right triangle is θ, then =.

3. (Yancheng, Jiangsu, 2008) Cut a piece of equilateral triangle paper along the height of one side, which can be spliced into quadrangles with different shapes. Try to write a quadrilateral name.

4. (Neijiang, Sichuan, 2008) As shown in the figure, in the rectangular grid diagram of, the number of rectangles excluding the shadow part is.

No answer

5.(2008 Foshan) As shown in the figure, it is known that P is a point on the diagonal BD of the square ABCD, and BP = BC, then ∠ACP degree is.

6. (Jiamusi City, 2008) In the following drawings, it is not the expanded diagram of the cube (fill in the serial number).

7. (Taian, 2008) If the sum of the upper and lower bottoms of the isosceles trapezoid is 4 and the acute angles of the two diagonals are 0, the area of the isosceles trapezoid is 0.

8. (Shaanxi Province, 2008) As shown in the figure, in the trapezoid,, and are considered as squares outside the trapezoid, and their areas are respectively, so the relationship is.

9. (Shaanxi Province in 2008) As shown in the figure, if the side length of the diamond is 2, the coordinates of the point are.

10, (Qingdao, Shandong, 2008) As shown in the figure, in the rectangular ABCD, diagonal AC and BD intersect at point O, if ∠ AOB = 60, AB = 4cm, the length of AC _ _ _ _ _ _ _ _ _ cm.

1 1, (Liangshan Prefecture, Sichuan Province, 2008) In the rhombus, the vertical bisector and vertical foot are. So, the area of the diamond is, and the length of the diagonal is.

12, (08 Hainan) As shown in the figure, if ABCD, AD‖BC, AE‖DC, AB=6cm in the isosceles trapezoid, AE= cm.

13, (Qinghai, 2008) It is known that the area of the rhombus is cm and the side length of the rhombus is cm; In an isosceles trapezoid,, cm, cm,, the waist length of the trapezoid is cm.

14, (Linyi, Shandong, 2008) As shown in the figure, in the rectangular ABCD, AB = 2, BC = 3, and the middle vertical line of diagonal AC intersects with AD and BC respectively.

Connect CE at point E and point F, and the length of CE is _ _ _ _ _ _ _ _.

15, (2008 Qiqihar) As shown in the figure, the side length of the diamond is1; Make a point, think about it, make a second diamond, and do it; Put forward a view that while making the third diamond, make it; By analogy, the side length of the first diamond is.

16, (Zhenjiang, Jiangsu, 2008) As shown in the figure, the side length of two congruent diamonds is 1 cm. An ant moves in a circle along the edge of the diamond in the order from this point, stops after walking 2008 cm, and then the ant stops at this point.

17, (Harbin, Heilongjiang, 2008) It is known that the side length of the rhombus ABCD is 6, the point E is on the straight line AD, DE = 3, and the connecting line BE intersects the diagonal AC at the point M, so the value is.

18, (Liangshan Prefecture, Sichuan Province, 2008) In the rhombus, the vertical bisector and vertical foot are. So, the area of the diamond is, and the length of the diagonal is.

19, (Yancheng, Jiangsu, 2008) If the midline length of the trapezoid is 3 and the height is 2, the area of the trapezoid is.

20. (Taiyuan, Shanxi, 2008) In the trapezoidal ABCD, AB=DC=3, and the trapezoidal ABCD is folded along the diagonal BD. If point A falls just at the midpoint E of the bottom BC, the circumference of the trapezoid is.

Fourth, the answer

First, multiple choice questions

1、5

2、0.6

3, parallelogram (or rectangle or diamond)

4. (Neijiang, Sichuan, 2008) As shown in the figure, in the rectangular grid diagram of, the number of rectangles excluding the shadow part is.

5、22.5

6、③

7. (The result is in the form of root symbol).

8、

9、

10、8cm

1 1、8

12、6

13、; four

14、

15、

16、

17,2 or

18、8

19、6

20、 15;

2 1、5

22、90

23、6

24. The answer is not unique. For reference: ① its internal angle is 60,60, 120,120; ② Its waist length is equal to the length of the upper sole; Its upper bottom is equal to half the length of its lower bottom.

25、 1

26、①②③④

27、

28、

29、 10㎝2

30、9

3 1、20

32、48

33, rectangle

34、

35、60

Third, answer questions.

1, solution: (1)BG=DE

∵ quadrilateral ABCD and quadrilateral CEFG are both squares,

∴GC=CE,BC=CD,∠BCG=∠DCE=90)

∴△BCG≌△DCE

∴BG=DE

(2) There are △ BCG and △DCE.

△BCG rotates 90 clockwise around point C, which coincides with△△ DCE.

2. Proof: In the square ABCD, take AB=2.

∫N is the midpoint of BC,

∴NC=

Yes,

And ∵NE=ND,

∴CE=NE-NC=，

,

Therefore, the rectangle DCEF is a golden rectangle.

3. Solution: within (1).

(2) Method 1: Connect the CD,

∫DE‖AC，DF‖BC，

∴ Quadrilateral DECF is a parallelogram,

Point d is the heart of △ABC,

∴ CD divides∠∠ ACB equally, that is ∠ FCD =∠ ECD.

∠ FDC =∠ ECD，∴∠ FCD =∠ FDC。

∴ FC=FD，

Decf is a diamond.

Evidence 2:

After D, DG⊥AB is in G, DH⊥BC is in H, and DI⊥AC is in i.

∫AD，BD ∠CAB，∠ABC，

∴DI=DG，

DG=DH。

∴DH=DI.

∫DE‖AC，DF‖BC，

∴ Quadrilateral DECF is a parallelogram,

∴S□DECF=CE DH =CF DI，

∴CE=CF.

∴□DECF is a diamond.

4.( 1) When E is the midpoint of CD, EB divides ∠AEC equally.

From ∠D=900, DE= 1, AD=, DEA=600, and in the same way ∠CEB=600, so ∠AEB=∠CEB=600, that is, EB shares ∠AEC equally.

(2)①∵CE‖BF,∴== ∴BF=2CE。

Ab = 2ce, ∴ point b bisects the line segment AF.

② Energy

Proof: cp =, CE= 1, ∠C=900, ∴EP=.

In Rt △ADE, AE= =2, AE = BF,

pb =，∴PB=PE

∠AEP=∠BP=900 ,∴△PAS≌△PFB。

∴△PAE can be obtained by rotating △PFB clockwise around point P.

The rotation degree is 1200.

5.( 1) quadrilateral BECF is a diamond.

It is proved that EF divides BC vertically,

∴BF=FC,BE=EC,∴∠ 1=∠2

∠∠ACB = 90°

∴∠ 1+∠4=90

∠3+∠2=90

∴∠3=∠4

∴EC=AE

∴BE=AE

CF = AE

∴BE=EC=CF=BF

∴ Quadrilateral BECF is a diamond.

(2) When ∠ a = 45. The diamond top is square.

Proof: ∫∠A = 45. ∠ACB=90 degrees.

∴∠ 1=45。

∴∠EBF=2∠A=90。

A diamond is square.

6.( 1) Prove that a quadrilateral is a rectangle,

(The diagonal of the rectangle is divided equally),

The opposite sides of a rectangle are parallel.

,.

(USA).

(2) At that time, the quadrangle was a diamond.

Prove that quadrilateral is rectangular,

The diagonal of the rectangle is divided equally.

Also obtained by (1),

,

A quadrilateral is a parallelogram (diagonal lines bisect each other).

The quadrilateral is a parallelogram)

Say it again,

A quadrilateral is a diamond (four parallel sides with diagonal lines perpendicular to each other).

The edge is diamond).

7.( 1) Proof: ∫AE‖BD, ∴∠ E = ∠ BDC.

DB division ∠ ADC ∴ ADC = 2 ∠ BDC

∫∠C = 2∠E

∴∠ADC=∠BCD

∴ trapezoid ABCD is an isosceles trapezoid.

(2) solution: from the problem (1), we get ∠ c = 2 ∠ e = 2 ∠ BDC = 60, BC = ad = 5.

∫In△BCD，∠ C = 60，∠ BDC = 30。

∴∠DBC=90

∴DC=2BC= 10

8. Solution: Solution: When cm, the area is;

When cm, the area is;

When cm, the area is.

(In each case, the graph gives 1, and the calculation result is correct, 1, ***6).

9. Prove that the intersection point C is CF⊥AB and the vertical foot is F. 。

∫ AB‖CD, AB∶CD, ∠ A = 90 in the trapezoid,

∴ ∠D=∠A=∠CFA=90。

The quadrilateral AFCD is a rectangle.

AD=CF，BF=AB-AF= 1。

In Rt△BCF,

CF2=BC2-BF2=8，

∴ CF=。

∴ AD=CF=。

E is the midpoint of AD,

∴ Germany =AE=AD=.

In Rt△ABE and Rt△DEC,

EB2=AE2+AB2=6，

EC2= DE2+CD2=3，

EB2+ EC2=9=BC2。

∴ ∠CEB=90。

∴ EB⊥EC.

10 and (1) prove that ∵ quadrilateral is a square, ∴ BC = CD, ∠ BCG = ∠ DCE = 90 .................. 2 points.

* ∴△ BCG ≌△ DCE. CG = CE .................. 4 points.

(2) Answer: The quadrilateral E'BGD is a parallelogram.

Reason: ∫△DCE rotates 90 clockwise around point D to get △DAE'

∴ce=ae′,∵cg=ce,∴cg=ae′,∵ab=cd,ab‖cd，

∴ DG, be DG, .................. 6 points.

∴ Quadrilateral is a parallelogram with .................. of 8 points.

1 1, solution 1: In the rectangle,,,

.

,,.

.

Solution 2: In a rectangle,

,,.

12, (1), that is, the quadrilateral is a parallelogram.

Divide equally,

Again,,,,

The quadrilateral is a diamond.

(2) Prove that 1: is the midpoint.

Again,,,

,)

,.

In other words, it is a right triangle. )

Evidence 2: Uniform, then, evenly distributed,

Let's begin.

Yes, the midpoint.

, is a right triangle. (7 points)

13, solution: (1) 36; (2) seconds;

(3) When three points form a right triangle, there are two situations:

(1) At that time, set a point to leave for one second.

Do sth.

,,.

At this time, the point left for a second.

(2) At that time, set a point to leave for one second,

,.

.

...

At this time, the point left for a second.

According to ① ②, when three points form a right triangle, the point leaves the point for seconds or seconds.

14, proof: (1),

From folding along the edge to overlapping, yes.

A quadrilateral is a rectangle with equal adjacent sides.

A quadrilateral is a square.

(2), and the quadrangle is a trapezoid.

A quadrilateral is a square.

Also points to the midpoint of. Connect.

In and,,,,

,.

A quadrilateral is a parallelogram.

...

The quadrilateral is an isosceles trapezoid.

Note: Event (2) can also be overdone, and the foot is the point of proof.

15, solution (1) proof: ∫ce bisection, ∴.

It's 2000 BC again, ∴.

In the same way.

(2) When the point o moves to the midpoint of AC, the quadrilateral AECF is rectangular.

Point o is the midpoint of AC. The quadrilateral AECF is a parallelogram.

Similarly, that is to say, the quadrilateral AECF is a rectangle.

The solution of 16 and (1) 1: as shown in Figure 25- 1.

A is AE⊥CD, and the vertical foot is E.

According to the meaning of the question, DE=.

In rt delta ade, AD=.

Scheme 2: as shown in Figure 25-2.

If point A is AE‖BC and point E is CD, then CE=AB=4.

∠AED=∠C=60。

∫∠d =∠c = 60，

∴△AED is an equilateral triangle.

∴AD=DE=9-4=5。

(2) Solution: as shown in Figure 25- 1.

CP = x, and h is the height on the side of PD. According to the meaning of the question, the area s of △PDQ can be expressed as:

S=PD h

=(9-x) x sin60

=(9x-x2)

=-(x-)2+。

We know from the meaning of the question that 0≤x≤5.

When x= (0≤x≤5), the maximum value of s =.

(3) Method 1: As shown in Figure 25-3.

Suppose there is a point m that satisfies the condition, and PD must be equal to DQ.

So 9-x=x= x, x=

At this time, the positions of point P and point Q are as shown in Figure 25-3, even if it is QP.

△PDQ is an equilateral triangle.

Q is QM‖DC, BC is M, and M is demand.

Connect MP and prove that the quadrilateral PDQM is a diamond.

Yi Zheng △ MCP △ QDP, ∴∠D=∠3.MP=PD.

∴MP‖QD, ∴ quadrilateral PDQM is a parallelogram.

And MP=PD, ∴ quadrilateral PDQM is a diamond.

So there is a point m that satisfies the condition, BM = BC-MC = 5-=.

Method 2: As shown in Figure 25-4.

Suppose there is a point m that satisfies the condition, and PD must be equal to DQ.

So 9-x=x= x, x=

At this time, the positions of point P and point Q are shown in Figure 25-4, and △PDQ is an equilateral triangle.

After point D, DO⊥PQ at point O extends the intersection of DO and BC at point M, connecting PM and QM, then DM vertically divides PQ, ∴ MP=MQ.

Yi Zhi ∠ 1 = ∠ C

∴PQ‖BC。

And ∵DO⊥PQ, ∴ MC ⊥ MD.

∴MP= CD=PD

That is MP=PD=DQ=QM.

∴ Quadrilateral PDQM is a diamond.

So there is a point m that satisfies the condition, BM = BC-MC = 5-=

17 and (1) proof,

.

Yes, the midpoint,

.

Say it again,

.

. )

,

.

Yes, the midpoint.

(2) Solution: The quadrilateral is a rectangle,

Prove this,

A quadrilateral is a parallelogram.

, yes, the midpoint,

.

Namely.

A quadrilateral is a rectangle.

18, (1) addition conditions: BE=DF or ∠BAE=∠DAF or ∠BAF=∠DAE, etc.

(2) Prove that the∵ quadrilateral ABCD is a diamond.

∴AB=AD

∠B =∠D

In △ Abe and ADF

AB=AD

∠B =∠D

BE=DF

∴△ABE≌ADF

∴AE=AF

19, the solution: (1) is in the parallelogram AB=CD. ，∠ A = ∠A=∠C，AD = CB，AB = CD。

E and f are the midpoint of AB and CD, respectively.

∴AE=CF

In and,

.

(2) If AD⊥BD, the quadrilateral BFDE is a diamond.

Prove,

Yes, it is the hypotenuse.

Yes, the midpoint,

.

As can be seen from the meaning of the question,

The quadrilateral is a parallelogram,

The quadrilateral is a diamond.

20. solution: ∵ quadrilateral ABCD is a square,

∴ AD=CD，∠A=∠DCF=900

And ∵DF⊥DE,

∴∠ 1+∠3=∠2+∠3

∴∠ 1=∠2

In Rt△DAE and Rt△DCE,

∠ 1=∠2

AD=CD

∠A=∠DCF

∴Rt△DAERt△DCE

∴DE=DF.

2 1, solution: (1) proof: ∫∠A = 90∠Abe = 30∠AEB = 60.

Eb = ed ∴∠ EBD = ∠ EDB = 30

∫pq‖BD ∴∠eqp=∠ebd∠EPQ =∠EDB

∴∠EPQ=∠EQP=30 ∴EQ=EP

The passing point e is EM⊥OP, and the vertical foot is ∴ pq = 2pm.m.

∫∠EPM = 30 ∴pm=pe ∴pe=pq

∫be = de = PD+PE ∴be=pd+ pq

(2) Solution: AE=BE ∴DE=BE=2AE.

∫ad = BC = 6 ∴ae=2 De =BE=4

When the point P is on the straight line ED (as shown in figure 1)

Do QH⊥AD at point Q = PQ = X.

PD=BE-PQ=4-x from ( 1)

∴y=PD QH=

When point P is on the extension line of line segment ED (as shown in Figure 2), point Q is QH⊥DA, and the extension line of DA is H' ∴ QH' = X.

Taking point E as EM'⊥PQ of point M' can also lead to EP=EQ=PQ ∴BE=PQ-PD.

∴PD=x-4 y=PD QH'=

(3) Solution: Connect PC and BD at point N (as shown in Figure 3) ∫ Point P is the midpoint of line segment ED.

∴ep=pd=2 ∴pq=∶DC = ab = AE tan 60 =

∴PC==4 ∴cos∠DPC== ∴∠DPC=60

∴∠QPC= 180 -∠EPQ-∠DPC=90

* pq‖BD ∴∠pnd=∠qpc=90 ∴pn=pd= 1

QC = =∠∠PGN = 90-∠FPC∠PCF = 90-∠FPC

∴∠ PCN =∠ PCF............... 1≈png =∠qpc = 90∴△png ~△qpc。

∴ ∴PG==

22, solution: when cm, the area is; When cm, the area is;

When cm, the area of is. (In each case, the number is given as 1, and the calculation result is correct 1, ***6).

23.( 1) Proof: At that time,

Say it again,

A quadrilateral is a parallelogram.

(2) Prove that quadrilateral is parallelogram,

.

.

(3) The quadrilateral can be a diamond.

Reason: As shown in the figure, connection,

As can be seen from (2),

Equal to each other.

At that time, the quadrangle was a diamond.

In,,

Once again,

,

Rotate clockwise around the point, and the quadrilateral is a diamond.

24, (1) method 1:

①∫ Quadrilateral ABCD is square, AC is diagonal,

∴ BC=DC，∠BCP=∠DCP=45。

PC = PC，

∴△PBC?△PDC(SAS)。

∴ PB= PD，∠PBC=∠PDC。

And ∵ PB= PE,

∴ PE=PD。

(2) (i) When the point E is on the line BC (E does not coincide with B and C),

PB = PE，

∴ ∠PBE=∠PEB，