In 2008, Xiamen students were classified according to the mathematical geometry questions of the national senior high school entrance examination.

(08 Henan Volume 18) 18. (9 points) When reviewing the knowledge of "congruent triangles", the teacher assigned an assignment: "As shown in Figure ①, it is known that in △ABC, AB=AC, P is any point in △ABC, and rotate AP clockwise around A to AQ, so ∠ QAP =.

BQ, CP, then BQ = CP. "

Xiao Liang is a thoughtful student. He proved △ abq △ ACP by analyzing Figure ①, so that after he proved BQ=CP, he moved the point P outside the isosceles triangle ABC and found that "BQ=CP" still holds. Please give a proof on Figure ②.

(08 Henan Province Volume 20) 20. (9 points) As shown in the figure, there is a river between A and B. Originally, from A to B, it needs to pass through DC and reach along the broken line A → D → C → B. Now a new bridge EF has been built, which can directly reach B from A along the straight line AB. All the way to BC= 1 1km,

∠ A = 45，∠ B = 37。 The DC bridge and the AB bridge are parallel, so how far is it from A to B now? (The result is accurate to 0. 1km ... Reference data: sin37 ≈0.60, cos37 ≈0.80).

(08 Henan Province Volume 2 1) 2 1. (9 o'clock) As shown in the figure, in the plane rectangular coordinate system, the coordinates of point A are (10,0), the coordinates of point B are (8,0), points C and D are on a semicircle M with a diameter of OA, and the quadrilateral OCDB is a parallelogram.

(08 Changsha, Hunan 19) 19. In the grid diagram below, the side length of each small square is 1 unit. Please draw the graph according to the following requirements:

(1) Draw the central symmetrical figure of the shaded part in Figure ① about the O point;

(2) Move the shaded part in Figure ② to the right by 9 units and draw a graph;

(3) Draw an axisymmetric figure of the shaded part with respect to the straight line AB in Figure ③.

(Figure ①) (Figure ②) (Figure ③)

(08 Changsha, Hunan 19) Sketch (2008), 2 points for each of the three parts, and 6 points for * * *.

(08 Changsha, Hunan, 24 questions) 24. (The full mark of this question is 8) As shown in the figure, in □ABCD, BC=2AB=4, and points E and F are the midpoint of BC and AD respectively.

(1) verification: △ Abe △ CDF;

(2) When the quadrilateral AECF is a diamond, find the area of the diamond.

(08 Changsha, Hunan, 24 questions) (1) proof; (4 points)

(2) When the quadrilateral AECF is a diamond, △ Abe is an equilateral triangle, (6 points)

The height of quadrilateral ABCD is, (7 points)

∴' s diamond origin is 2. (8 points)

(08 Hunan Changde 2 1) 2 1. As shown in Figure 4, it is known that ⊙O is the circumscribed circle of △ABC, and AB is the diameter. If PA⊥AB and P0 pass through the midpoint m of AC,

Prove: PC is the tangent of ⊙ O.

(08 Changde, Hunan 2 1 Question Answers) Proof: Connect OC,

∵PA⊥AB, ∴∠PA0=900, ......................... 1 min.

∫PO passes through the midpoint m of AC, OA=OC,

∴PO bisects ∞∠AOC,

∴∠ AOP = ∠ COP ................................. 3 points.

∴ in △PAO and △PCO, there are

OA=OC，∠AOP=∠COP，PO=PO，

∴△pao?△pco, ....................... scored 6 points.

∴∠PCO=∠PA0=900，

That is, PC is 7 points of tangent ..................... of ⊙ O.

23. As shown in Figure 7, in the trapezoidal ABCD, if AB//DC, AD=BC, diagonal BD and AC divide the trapezoid into four small triangles.

(1) List all possible situations of selecting two triangles from these four small triangles, and find out the probability that the two triangles selected are similar triangles (note: congruence is regarded as a special case of similarity)?

Please select a group of similar triangles and give proof.

Solution (1)

(08 Hunan Changde Question 23 Solution) Solution: (1) All possible situations of two triangles are as follows:

1②, 1③, 1④, 2③, 2④, 3④ .................................................................................. scored 2 points.

The two groups (① ③, ② ④) were similar.

The probability that the two selected triangles are similar triangles is p =.

(2) Proof: choose ① and ③ to prove.

In △AOB and △COD, ∫AB‖CD,

∴∠CDB=∠DBA，DCA=∠CAB，

∴△AOB∽△cod 8 points.

Choose ② and ④ to prove it.

∵ Quadrilateral ABCD is an isosceles trapezoid, ∴∠ DAB = ∠ cab.

∴ in △DAB and △CBA, there are both.

AD=BC，∠DAB=∠CAB，AB=AB，

∴△dab?△CBA, 6 points.

∴∠ADO=∠BCO.

And ∠ DOA = ∠ COB, ∴△ DOA ∽△ COB ....................................... 8 points.

(08 Hunan Changde 26 questions) 26. As shown in Figure 9, there are △ABC and right-angled trapezoid DEFG on the straight line, with CD = 6 ㎝; In △ABC: ∠ C = 90o, ∠ A = 300, AB = 4 ㎝; In the right-angled trapezoid DEFG: EF//DG, ∠ DGF = 90o, DG = 6 ㎝, DE = 4 ㎝, ∠ EDG = 600. Answer the following questions:

(1) rotation: rotate △ABC 900 clockwise around point C. Please draw the corresponding figure in the drawing after rotation.

△A 1B 1C, and find the length of AB 1;

(2) Folding: Fold △A 1B 1C along a straight line that passes through point B 1 and is perpendicular to the straight line, so as to obtain the corresponding graph after folding.

△A2B 1C 1, try to determine the shape of quadrilateral A2B 1DE? And explain the reasons;

(3) Translation: Translate △A2B 1C 1 to the right along a straight line to △A3B2C2. If the translation distance is x, the overlapping area of △A3B2C2 and the right trapezoid is y, and when y is equal to half of the area of △ABC, what is the value of x?

(08 Hunan Changde 26 questions)

Solution: (1) is given by △ABC: BC=2, AC = AB× COS 30 =,

∴ AB1= AC+CB1= AC+CB = ................................................................................... 2 points.

(2) The quadrilateral A2B 1DE is a parallelogram for the following reasons:

∠∠EDG = 60，∠a2b 1c 1 =∠a 1b 1c =∠ABC = 60 ,∴a2b 1‖de

And a2b1= a1b1= ab = 4, de = 4, ∴ a2b 1 = de, so the conclusion holds. ...............................................................................................

(3) According to the meaning of the question:

S△ABC=，

(1) when or, y = 0.

At this time, the area of the overlapping part will not be equal to half of the area of △ABC.

② If the overlapping length of the right-angle side B2C2 of the isosceles trapezoid and the bottom DG is DC2 = c1C2-DC1= (x-2) ÷, then y =,

When y= S△ABC=, that is,

Solve (give up) or.

Where appropriate, the area of the overlapping part is equal to half the area of △ABC.

③ When △A3B2C2 and isosceles trapezoid completely coincide, that is, 7 o'clock.

④ When appropriate, B2G = B2C2-GC2 = 2-(-8) = 10-

Y =，

When y= S△ABC=, that is,

Solve, or (give up).

∴ Where appropriate, the area of overlapping part is equal to half the area of △ABC. ...................................................................................................................................................

From the above discussion, it can be seen that when or, the area of the overlapping part is equal to half of the area of △ABC. ..............................................................................................................................................

(08 Chenzhou, Hunan 2 1 Title) 2 1 Drawing Title:

As shown in fig. 6, firstly, ABC is moved down by 4 units, and then the reflection of the symmetry axis is carried out with the straight line L as the symmetry axis. Please sum in the given grid paper in turn.

(08 Hunan Chenzhou 2 1) The drawing is correct, with 3 points each (omitted) and 6 points.

(Hunan Chenzhou 22 questions) 22. After the Wenchuan earthquake, the rescue team sent helicopters to rescue villages A and B. The plane was at point P, 450 meters above the ground, and the depression angle of village A was measured as (. As shown in fig. 7). Find the distance between village A and village B (the result is accurate to the meter, refer to the data).

(08 Chenzhou, Hunan, 22 questions). Solution: According to the meaning of the problem,

So, so,

So AB=PB 3 points

At,, PC=450,

So PB = 5 points

So (m)

Answer: A little .6 points.

As shown in Figure 8, ABC is an isosceles triangle. After folding along the bottom BC, δDBC can be obtained. Please judge the shape of the quadrilateral ABDC and give your reasons.

The quadrilateral ABCD is a diamond with two points.

The reason is:

You get △ ABC △ DBC by folding. So four points.

Because △ABC is an isosceles triangle,

therefore

So AC = CD = AB = BD, 7 points.

So the quadrilateral ABCD is a diamond with 8 points.

Note: If students only answer that the quadrilateral ABCD is a parallelogram, give 1 point, and if the reasoning is correct, give 5 points and ***6 points.

(08 Huaihua, Hunan, 24 questions) 24. (The full score of this question is 7 points)

As shown in figure 10, quadrilateral ABCD and DEFG are both squares, connecting AE and CG, AE and CG intersect at point m, CG and AD intersect at point n. 。

Verification: (1);

(2)

It is proved that (1) quadrilateral and quadrilateral are both squares.

3 points

4 points

(2) By (1)

7 points

∴ AMN∽ CDN 6 points

(08 Huaihua, Hunan, 25 questions) 25. (The full score of this question is 7 points)

As shown in figure 1 1, it is known that the area of △ is 3 and AB=AC. Now △ is translated in the direction of CA to get△.

(1) Find the area of the quadrilateral CEFB;

(2) Try to judge the positional relationship between AF and BE, and explain the reasons;

(3) If, find the length of AC.

Solution: (1) comes from the essence of translation.

.3 points

(2) The proof is as follows: (1) The quadrilateral is a parallelogram.

5 points

(08 Huaihua, Hunan, 26 questions) 26. (The full score of this question is 7 points)

There is a soil slope next to the teaching building of a school, and the soil slope is a flat land, as shown in figure 12. The slope is long and has a gradient. In order to prevent landslides and ensure safety, the school decided to transform the soil slope, and geologists surveyed it to ensure that the mountain would not landslide when the slope angle was less than 0.

(1) Find the length of the vertical distance from B slope to the ground before reconstruction;

(2) In order to ensure safety, the school plans to keep the foot of the slope motionless and the edge of the top of the slope cut in. How many meters at least?

(08 Hunan Huaihua 26 questions)

(08 Hunan Xiangtan 18) 18. (The full mark of this question is 6 points)

As shown in the figure, each small square in the grid paper is a square with a side length of 1 unit. After the plane rectangular coordinate system is established, the vertex of ABC is on the grid point, and the coordinate of point B is (5, -4). Please make it symmetrical about y and ABC, and write down the coordinates.

(08 Hunan Xiangtan 18 Question Answers) Cartography (omitted) 4 points.

The coordinates of the point are (-5, -4) 6 points.

(08 Xiangtan, Hunan, 20 questions) 20. (The full mark of this question is 6 points)

As shown in the figure, the quadrilateral ABCD is a rectangle, e is a point on AB, and DE=AB, C is CF⊥DE, and the vertical foot is F. 。

(1) conjecture: the relationship between AD and CF;

Please prove the above conclusion.

(08 Hunan Xiangtan 20 questions) Solution: (1) .2 points.

(2) The quadrilateral is a rectangle,

3 points

Another 4 points

5 points

6 points

(08 Hunan Xiangtan 24 questions) 24. (8 points for this question)

As shown in the figure, the diameter of AB =4, point P is a point on the extension line of AB, and the tangent line passing through point P is C, connecting AC.

(1) If ∠ CPA = 30, find the length of PC;

(2) If point P moves on the extension line of AB, the bisector of ∠CPA intersects with AC at point M. Do you think the size of ∠CMP has changed? If yes, please explain the reasons; If not, find the size of ∠CMP.

(08 Hunan Xiangtan 24 questions) Solution: (1) Link OC,

Yes, the tangent,

4 points

(2) The size has not changed by 5 points.

6 points

7 points

8 points

(08 Yiyang, Hunan 18) 18. As shown in Figure 8, in △ABC, AB=BC= 12cm, ∠ABC = 80, BD is the bisector of ∠ABC, and de ∠ BC.

(1) Find the degree of ∠EDB;

(2) Find the length of DE.

(08 Hunan Yiyang 18 problem solution) solution: (1)∫DE‖BC,

∴∠ EDB =∠ DBC = 3 points

(2) ∵ AB = BC, BD is the bisector of ∠ABC, and ∴D is the midpoint of AC.

In \de \u BC, \ e is the midpoint of AB,

∴ Germany = 6 points

(08 Yiyang, Hunan, Question 22) 22. △ABC is an equilateral triangle scrap iron, which is used to cut a square deFG, so that one side of the square DE falls on BC, and the vertices F and G fall on AC and AB respectively.

I. Proof: △ BDG △ CEF;

Two. Inquiry: How to accurately draw a square on the iron sheet?

Xiao Cong and Xiao Ming each made an idea. Please choose one of the two questions IIA and IIA that you like to answer. If you solve both questions, you will only score with the answer of IIA.

Ⅱ A. Xiao Cong thinks: To draw a square DEFG, as long as the side length of the square can be calculated, the lengths of BD and CE can be calculated, and then the points D and E can be determined, so it is easy to draw a square DEFG.

Let the side length of △ABC be 2, please help Xiao Cong find the side length of the square (the result is expressed by a formula with a root sign, and the denominator does not require a rational number).

Ii B. Xiao Ming thinks that you can draw a square without finding the side length of the square. The specific method is:

① Take any point G' on the side of AB, as shown in the figure, as a square G' d' e' f ";

② Connect BF' and extend AC to F;

③ If FE‖F'E' crosses BC in E, FG‖F'G' crosses AB in G, and GD‖G'D' crosses BC in D, then the quadrilateral DEFG is the demand.

Do you think Xiao Ming's approach is correct? Explain why.

(08 Yiyang, Hunan, 22 questions) 1. Proof: ∵DEFG is a square.

∴GD=FE, ∠ GDB = ∠ FEC = 90 2 points.

∫△ABC is an equilateral triangle with∴∠ B =∠ C = 60 3 points.

∴△BDG≌△CEF(AAS) 5 points

2 a. solution 1: let the side length of the square be x and the height AH be △ABC.

Get 7 points

From △AGF∽△ABC: 9 points.

Solution: (or) 10 score.

Solution 2: Let the side length of the square be x, and then score 7 points.

In Rt△BDG, tan∠B=,

9 points

Solution: (or) 10 score.

Solution 3: Let the side length of a square be x,

Then 7 points.

From Pythagorean Theorem: 9 points

Solution: 10 score

B. Solution: Correct 6 points

It is known that the quadrilateral GDEF is a rectangle with 7 points.

∫FE‖F ' e '，

∴ ,

Similarly,

∴

And F 'E' = F 'G',

∴FE=FG

Therefore, a rectangular GDEF is a square 10 point.

(08 Yiyang, Hunan, Question 23) 23. Two congruent right triangles ABC and DEF overlap, where ∠ A = 60 and AC= 1. Fix △ABC and do the following for △DEF:

(1) As shown in figure 1 1( 1), △DEF moves to the right along the line segment AB (that is, the point D in the line segment AB moves), connecting DC, CF and FB, and the shape of quadrilateral CDBF keeps changing, but its area remains unchanged, so we request its area.

(2) As shown in figure 1 1(2), when point D moves to the midpoint of AB, please guess the shape of the quadrilateral CDBF and explain the reason.

(3) As shown in figure 1 1(3), fix the point D of △DEF at the midpoint of AB, and then rotate △DEF clockwise around the point D to make DF fall on the side of AB. At this time, point f coincides with point b, and AE is connected. Please find the value of sinα.

(08 Hunan Yiyang 23 problem solution) Solution: (1) CG⊥AB is after C in G.

At Rt△AGC, sin 60 =, ∴ 1 point.

∵AB=2, ∴S trapezoidal CDBF=S△ABC= 3 points.

(2) Diamond 4 points

Cdbf, FC‖BD, ∴ Quadrilateral CDBF is a parallelogram with five points.

∫df‖AC, ∠ ACD = 90, ∴CB⊥DF 6 points.

∴ Quadrilateral CDBF is a diamond with 7 points.

(Judging that the quadrilateral CDBF is a parallelogram and proved to be correct, score 2 points)

(3) Solution 1: If DH⊥AE is at H after the intersection D, then S△ADE= 8 points.

S△ADE=, 9 points

∴, sinα= 10 in Rt△DHE'.

Solution 2: ∫△ADH∽△ABE 8 points.

∴

Namely:

9 points

∴sinα= 10 integral

(08 Yongzhou, Hunan 19) 19. (6 points) As shown in the figure, the side length of each square in the left grid paper is A, and the side length of each square in the right grid paper is B. Rotate the figure in the left grid paper clockwise by 90 and draw it in the right grid paper according to the ratio of B: A. 。

(08 Yongzhou, Hunan 19) (6 points)

(Hunan Yongzhou 22 questions) 22. (8 points) As shown in the figure, △ABC and △CDE are equilateral triangles, and points E and F are on AC and BC respectively, and EF‖AB.

(1) Verification: the quadrilateral EFCD is a diamond;

(2) Let CD = 4 and find the distance between point D and point F. 。

(08 Answers to 22 Questions in Yongzhou, Hunan) (8 points)

(1) proves that sum is an equilateral triangle.

1 point

2 points

and

3 points

The quadrilateral is a diamond with four points.

(2) Solution: Links intersect at point 5.

From this, we can see 6 o'clock.

7 points

8 points

(08 Yongzhou, Hunan, 24 questions) 24. (10 point) As shown in the figure, it is known that the diameter of ⊙O is AB = 2, the straight lines M and ⊙O are tangent to point A, P is the moving point on ⊙O (not coincident with points A and B), and the extension line of PO intersects with ⊙O at point C.

(1) Verification: △ APC ∽△ COD.

(2) Let AP = x and OD = y, and try to express y with an algebraic expression containing x. 。

(3) When trying to find out what the value of x is, △ACD is an equilateral triangle.