First of all, the conditions are open and exploratory.

Give the conclusion of the problem, let the problem solver analyze and explore the conditions that should be met to make the conclusion hold, but the conditions that meet the conclusion are often not unique. Such a question is an open question. It requires the problem solver to be good at starting from the conclusion of the problem, pursuing in reverse and seeking in many ways. This kind of questions are often designed with the basic knowledge as the background, mainly to examine the ability of the problem solver to master and summarize the basic knowledge.

Example 1, (Yu Xi, 2005). As shown in fig. 8, it is known that AB = AD, ∠ 1 = ∠ 2. In order to make △ ABC △ ade, the conditions that need to be added are (only fill in one).

Solution: ∠ B = ∠B=∠D or ∠ C = ∠C=∠E or AC = AE.

Example 2, (Changsha, 2005). As shown in the figure, AB = AC,

The condition to be added is _ _ _ _ _ _ (only one condition is added).

Solution: AD=AE or ∠B=∠C or ∠ADC=∠AEB.

Example 3, (Jinhua, 2005)

As shown in the figure, in △ABC, point D is on AB, point E is on BC, and BD = Be.

(1) Please add another condition to make △ BEA △ BDC and give the proof.

The conditions you added are: _ _ _ _ _ _ _ _

(2) According to the conditions you added, write a pair of congruent triangles in the picture: _ _ _ _ _ (just a pair of congruent triangles, no other line segments, no other letters, no need to write the proof process).

Tip: (1)≈BAE =∠BCD or ∠AEB=∠CDB or AE=CD, the proof is abbreviated; (2)△ADC?△AEC

Example 4 (Fuzhou Curriculum Reform Volume in 2005)

Known: As shown in Figure 7, point C and point D are on the AB line, and PC = PD.

Please add a condition to make congruent triangles exist in the diagram and prove it.

The added condition is _ _ _ _ _, and you get a pair of congruent triangles that are △ _ _ △ _ _ _.

Prompt: The added conditions are: ∠ A = ∠A=∠B (or PA = Pb or AC = BD or AD = BC or APC = ∠ BPD or APD = ∠ BPC, etc. ).

Congruent triangles is △ PAC △ PBD (or △ APD △ BPC).

Proof: (omitted)

Second, the conclusion is open and exploratory.

Given the conditions of the problem, the problem solver can explore the corresponding conclusions according to the conditions, and the conclusions that meet the conditions often present diversity, or the "existence" of the corresponding conclusions requires the problem solver to infer the situation, and even needs the problem solver to explore the conclusions of changing conditions. These questions are all open-ended, and the problem solver needs to make full use of the conditions to make bold and reasonable guesses, discover the laws and draw conclusions. This kind of problem mainly examines the divergent thinking of the problem solver and the application ability of the basic knowledge learned.

Example 5 (Anhui in 2005). As shown in the figure, AB ‖ DE, AB = DE and AF = DC are known. How many pairs of congruent triangles are there in the picture? Choose one of them to prove it.

Solution: There are three pairs of congruent triangles in the picture, namely △ ABF △ DEC.

△ABC?△DEF，△BCF?△EFC .

Proof: ∫ab‖de, ∴∠ A = ∠ D.

AB = DE，AF=DC，

∴△ABF≌△DEC。

Example 6 (Ningbo, 2005). As shown in the figure, in △ABC, AB=AC, the intersection point A is GE‖BC, the angular bisectors BD and CF intersect at H, and their extension lines intersect at E and G respectively. Try to find three pairs of congruent triangles in the picture and prove one pair of congruent triangles.

Tip: △ AGC △ AFB. △ AGF △ DFD. △ HBF △ HDC. △ AFC △ ADB. Brief proof

Example 7. (Changzhou in 2005)

As shown in the figure, it is called an equilateral triangle,

They are equilateral triangles on the sides and at the top.

(1) Besides the known equilateral, please guess which equilateral line segments are there.

And prove that your guess is correct;

(2) You have proved that how can equal line segments get each other?

Write the process of change.

Tip: (1) AE = BF = CD; AF = BD = CE Proof: (omitted)

(2) Rotate around E, D and F, and then fold in half.

Example 8. (ponytail in 2005) Two congruent equilateral triangles △ABC and △ACD are combined to form a diamond ABCD. Superimpose a 60-degree triangular ruler on this diamond, so that the vertex of the 60-degree angle of the triangular ruler coincides with point A, and the two sides coincide with AB and AC respectively. Turn the triangular ruler counterclockwise around point A. 。

(1) What conclusions can BE drawn by observing or measuring the lengths of Be and CF when the two sides of the triangular ruler intersect with the two sides of the diamond BC and CD at points E and F respectively (as shown in figure 13- 1)? And prove your conclusion;

(2) When the two sides of the triangular ruler intersect with the extension lines BC and CD on both sides of the diamond at points E and F respectively (as shown in figure 13-2), is the conclusion you got in (1) still valid? Briefly explain why.

Solution: (1) be = cf.

It is proved that in △ABE and △ACF, ∫∠BAE+∠EAC =∠CAF+∠EAC = 60,

∴∠BAE=∠CAF.

ab = ac，∠B=∠ACF=60 ,∴△ABE≌△ACF(ASA).

∴BE=CF.

(2)BE=CF still holds. According to the triangle congruence axiom, △ABE and △ACF can also be proved.

Example 9. As shown in the figure, A, B, C and D are on the same straight line, AB = CD, DE‖AF and DE = AF. Proof: △ AFC △ Deb. If BD moves in parallel along the direction of AD side, as shown in the figure, when point B coincides with point C, as shown in the figure, when point B is on the right side of point C, other conditions remain unchanged. If not, please explain why.

Proof: ∫de af, ∴∠ A = ∠ D,

Ab = cd, ∴ AB+BC = CD+BC, that is, AC = DB.

In △AFC and △DEB,

AC = DB，∠A=∠D，AF=DE，

∴△AFC≌△DEB.

Example 1 1. As shown in figure (1), AB⊥BD, ED⊥BD, AB = CD, BC = DE are known, and the verification is AC ⊥ CE. If the CD is translated in the CB direction, as shown in Figure (2), (3) and (4). Please explain the reason.

Tip: You can prove △ ABC△ CDE and get ∠ ACB = ∠ E.

∠∠ACB+∠ECD =∠E+∠ECD = 90，

∴∠ACE= 180 -90 =90 ,∴AC⊥CE.

Figures (2), (3), (4) and (5) show that AC 1⊥C2E still holds, which proves the same as above.

Example 12. As shown in figure (1), in △ABC, ∠ BAC = 90, AB = AC, AE is a straight line passing through A, B and C are on opposite sides of AE, BD⊥AE is in D, and CE⊥AE is in E, which proves: (. (2) If the straight line AE rotates around point A to position (2) (BD < CE), other conditions remain unchanged. What is the relationship between BD and DE and CE? Please prove it. (3) If the straight line AE rotates around point A to the position in Figure (3), and (BD > CE) other conditions remain unchanged, what is the relationship between BD and DE and CE? Please write the result directly without proof. (4) Summary (1), (2) and (3). Please express the relationship among BD, DE and CE in simple language.

Proof: (1)∵BD⊥AE, CE⊥AE (known), ∴ BDA = ∠ AEC = 90 (vertical definition).

∠∠BAD+∠CAE = 90，∠BAD+∠ABD=90，

∴∠CAE =∞∠Abd (complementary angles of the same angle are equal)

In △ABD and △CAE

∴△ABD≌△CAE(AAS)，

∴ BD = AE, AD = CE (the corresponding sides of congruent triangles are equal).

∵AE=AD+DE,∴AE=CE+DE，

∴BD=CE+DE.

(2) BD = de-ce, and the proof method is the same as (1).

(3)BD=DE-CE。

(4) Induction (1)(2)(3) The conclusion can be expressed as:

When b and c are opposite to AE, BD = de+ce; When b and c are on the same side of AE, BD = de-ce;

Note: The key to this problem is to guess the law of quantity in dynamic geometry and then prove it with geometric knowledge.

22. As shown in the figure, in the square grid of 10×5, the side length of each small square is 1. Translate △ABC to the right by 4 units to get △A'B'C', and then rotate △A'B'C' counterclockwise by 90.

22. Draw A A'B'c' correctly as shown in the picture.

Draw △‘B'C b' c correctly.

(Note: The position of A A'B'C is incorrect, but the correct △A' B' C is drawn on the basis of △'B'C, and 3 points are scored. )

Three: strategic opening and exploration

The open strategy problem generally refers to the problem that the problem solver can't find the unique solution or the solution path is unclear. This kind of problem requires the problem solver to be unconventional, be good at innovation, pursue multiple solutions to one problem, and give the problem solver a broad thinking space. Through active thinking, innovative exploration, exploration of problem-solving strategies and ideas, flexible use of problem-solving ideas and methods, optimization of problem-solving programs and processes.

Example13 (Shiyan Curriculum Reform Volume in 2005) is shown in the figure, and △ABC is known. Please add a condition and write a conclusion to prove your conclusion.

The additional conditions are:

Known:

Verification:

Proof: The added condition is BD=CE. The conclusion is ∠ b = ∠ c.

It is proved that in Rt△BEC and Rt△CDB,

BD = CE BC = BC

∴Rt△BEC≌Rt△CDB。

∴∠B=∠C

Example 14. (Yangzhou, 2005) As shown in the figure, in △ABC and △DEF, D, E, C and F are on the same straight line, and there are four conditions below. Please choose three of them as topics and the rest 1 as conclusions, write a true proposition and prove it.

①AB=DE，②AC=DF，③∠ABC=∠DEF，④BE=CF .

Known:

Verification:

Prove:

Tip: The answer is not unique, such as known: ① ② ④; Verification: ③ or known:13④; Verification: 2.

24 (Zhangzhou, 2005). As shown in the figure, five equivalent relationships are given: ①AD=BC, ②AC=BD, ③CE=DE, ④∠D=∠C,

⑤ DAB = ∠ CBA. Please write a correct proposition (only write one case) with two of them as conditions and one of the other three as conclusion, and prove it.

27. (9 points for this question)

As shown in the figure, in the quadrilateral ABCD, point E is on the edge CD, AE and Be are connected, and the following five relationships are given: ① AD ‖ BC; ②DE = CE； ③∠ 1=∠2; ④∠3=∠4; ⑤ AD+BC = AB。 Take three of them as the topic and the other two as the conclusion to form a proposition. (1) Write a true proposition with serial number (writing form: if ××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××

(2) Write three true propositions with serial numbers (without proof);

(3) Extra points: there are more than four true propositions. When you think about it, you can write more photo questions. For each real question, you can write 1 point, with a maximum of 2 points.

27. Solution: (1) If 123, then 45.

Proof: as shown in the figure, the extension line runs from AE to BC to F.

∴∠ BC1= ∠ f

* aed = cef，DE=EC∴△ADE≌△FCE.

∴AD=CF,AE=EF

∠∠l =∠F，∠ 1=∠2。 ∠2=∠F

∴AB=BF∴∠3=∠4

∴AD+BC=CF+BC=BF=AB

(Note: The proofs of other true propositions can be graded according to the above process. )

(2) If ① ② ④, ③ ⑤

If ① ③ ④, then ② ⑤

If it is 13 15, then it is 24.

(3) If the four propositions in (1)(2) contain false propositions ("If 234, then 1 15"), no extra points will be given; If (3) contains false propositions, no points will be added.

2 1- (the full mark of this question is 8) As shown in the figure, of the following four conditions, please take two of them as known conditions and the third as the conclusion, and deduce a correct proposition (write only one case).

2 1. Proof: condition1AE = ADAB = AC2AB = AC ∠ B = ∠ C3AE = AD ∠ B = ∠ C.

Example 15 is shown in the figure. It is known that AD = BC, AB = DC and DE = BF. Try to explore: are BE and DF equal? .

Analysis: To prove BE = DF, we must prove △ Abe △ CDF, and prove that these two triangles are the same. Two sets of conditions are met, AB = CD. AD+DE = CB+BF, that is, AE = CF, and then prove ∠ A = ∠ A = ∠ C. Then observe ∠ A and C.

It is obvious from the conditions of AD = CB and AB = CD that if BD is connected, then the conditions of △ AB=CD and △CDB have been met, and the conclusion can be proved.

Solution: Equality. Reason:

The link BD is at △ABD and △CDB.

∴△ABD≌△CDB(SSS)

∴∠ A =∠ C (the corresponding angles of congruent triangles are equal).

Ad = cb, DE = BF (known), ∴ AD+DE = CB+BF, that is, AE = CF

In △ABE and △CDF.

∴△ABE≌△CDF(SAS).

∴ be = df (the corresponding sides of congruent triangles are equal).

Note: (1) When solving related problems, we often encounter situations where the known conditions and conclusions cannot be communicated. At this time, it is necessary to add auxiliary lines to create conditions to serve the conclusion. (2) Using congruent triangles to prove that the line segments are equal or the angles are equal, we often need to add auxiliary lines to construct triangles. There are two situations when constructing: ① The line segment or angle to be proved is not in two triangles that may be the same on the graph. (2) Some qualified congruent triangles can't be directly displayed in the graph, so it needs to be found by adding auxiliary lines, such as △ABD and △ CDB in this question.

Example 16. Known: As shown in the figure, AB = AC, DB = DC.

(1) If e, f, g and h are the midpoint of each side, then verify: eh = fg.

(2) If the links AD and BC intersect at point P, what is the relationship between AD and BC? Prove your conclusion.

Solution: (1) Proof: Linked advertisement.

In △ABD and △ACD,

∴△ABD≌△ACD,∴∠ABD=∠ACD.

At △BEH and △CFG,

∴△BEH≌△CFG

∴EH=FG.

(2)AD is perpendicular to BC and bisects BC,

Let AD and BC meet at p.

∠ BAP =∠ CAP from ( 1)，

Yi Zheng △ BAP △ Cap, ∴ Pb = PC, ∠ APB = ∠ APC,

∠ APB +∠ APC = 180，

∴∠ APB = 90, so AD⊥BC and AD share BC equally.

Note: (1) congruent triangles can not only get equilateral and equilateral, but also further deduce some properties of the graph according to equilateral and equilateral, such as two straight lines are parallel and two straight lines are vertical. In this case, the first congruence provides the condition of the third congruence. This shows the instrumental role of congruent triangles's knowledge.

(2) Through the previous research, we can see that in congruent triangles's proof related issues, the following two basic graphics are often involved:

The first category is about angles. As shown in the figure, the same feature of these three figures is that a set of angles corresponding to two triangles have "common * * * parts".

The second category is about edges, as shown in the figure.

The common feature of these three figures is that a set of corresponding sides of two triangles have "common parts".

Mastering the characteristics of these basic graphics, separating them from more complex graphics and making full use of the relationship between the sides or angles of common * * * will help us find a quick proof method.

Example 18. After a triangular glass in the greenhouse was destroyed, only the shadow part of the figure was left. After you measure the data in the picture, you can cut the triangle glass that meets the specifications in the building materials sales department and explain the reasons.

Tip: Measure ∠ABC, ∠DCB and line BC. The two angles and sides determine the shape and size of the triangle.

Example 19. As shown in the figure, it is known that in △ABC, AB=AC, ∠ BAC = 90, and the straight lines passing through B and C respectively are vertical, and the vertical feet are E and F.

(1) It is proved that if the straight line passing through A does not intersect the hypotenuse BC, there is EF=BE+CF, as shown in figure 1.

(2) As shown in Figure 2, when the straight line passing through A intersects the hypotenuse BC, other conditions remain unchanged. What conclusion can you draw? Please prove it.

19.( 1) certificate △ BAE △ caf; (2)EF=BE-CF .

Example 19 it is known that the outer diameter of the part is a, and to calculate its thickness x, it is necessary to first calculate the inner hole diameter AB, make a simple tool, and use the triangle congruence to calculate AB.

Pointing: For AB, it is the diameter of the inner hole, which cannot be directly measured, and it is not easy to make it vertical, so SAS can be used to take the midpoint, which reminds people of scissors and pliers. This method can be used to measure AB, as shown in the figure.

Solution: You can design a pliers-like tool as shown in Figure 5-70. The length of CD is the distance between A and B. 。

AB=a-2x。

Four: scene opening and exploration

Given the actual situation of the problem, the problem solver is required to establish a mathematical model, seek practical methods to solve practical problems, or use mathematics to design various schemes to provide decision-making basis. This kind of problem is called situational openness, which is often based on the actual situation or real life and involves social production, science and technology, economy and mathematics itself. The answer to this kind of question is innovation itself, so that students can develop their awareness of applied mathematics in creation.

As shown in Figure 20, A and B are located at two ends of a pond. Xiaoli wants to measure the distance between A and B with a rope, but the rope is not long enough. Can you help her design the measurement scheme? If not, explain where the difficulty lies; If you can, write a plan and explain the reasons.

Feeling: If you find a rope long enough, you can measure it directly. If there is not enough rope, we will construct a congruent triangles on the lakeshore and "move" AB to the land for measurement. Just measure the short rope several times.

Solution 1: Yes.

Measurement scheme: (1) First, take a point on the land that can directly reach point A and point B;

(2) Connect AC and extend to point D, so that CD = CA.

(3) Connect BC and extend to point E so that CE = CB.

(4) Connect DE and measure its length.

As shown in Figure 5- 105, the length of DE is the distance between A and B. 。

Reason: in △ABC and △DCE.

∴△ABC?△DCE(SAS)。

Company.

Solution 2: Yes.

Measurement scheme: (1) Take two points C and D on the vertical AF of AB, so that CD = AC.

(2) Take point D as the vertical DG of AF, and take point E on DG to make points B, C and E on the same straight line;

(3) The measured length DE is the distance between A and B, as shown in the figure.

Reason: links b, c, e,

∵ Points B, C and E are on the same straight line,

∴ ∠ 1=∠2,

* ab⊥af,dg⊥af，

∴ ∠BAC=90 =∠GDC。

In △ABC and △DEC,

∴△ABC?△dec(asa)。

Company.

Solution 3: Yes.

Measurement scheme: (1) Send a classmate to wear a sun hat and stand at attention at point A;

(2) Ask students to adjust their hats so that their eyes pass through the "brim" and fall on point B across the lake;

(3) The classmate turned an angle and kept the posture just now, and the "brim" did not move. At this time, he looked out and still let his eyes pass through the "brim", and the position where his eyes fell was point C;

(4) Connect AC and measure the length of AC, that is, the distance between A and B, as shown in the figure, which is a schematic side view.

Reason: According to the measurement: ∠ BDA = ∠ CDA.

* da⊥bc，

∴ ∠DAB=∠DAC=90。

In △ADB and △ADC.

∴△ ADB△ ?△ADC(asa).

∴ AB=AC。

Guidance: There is often more than one solution to the practical problems in life. When choosing a specific method, we should consider the specific situation. Similarly, we use triangle congruence to measure distance. Scheme 3 is relatively simple, but we have to repeat it 2-3 times, and then take the average value to avoid a big error.

Example 2 1 in the process of building a railway, a railway construction team needs to get through a hill, and the length of the tunnel should be measured during design. The front of the mountain is just a clearing. Under this favorable terrain, can surveyors use the knowledge of triangle congruence to measure the length of the tunnel to be excavated? Tell the truth.

Pointing: it is difficult to directly measure a and B. Therefore, we can use the open space in front of the mountain to construct two congruent triangles, make a pair of sides corresponding to AB equal, and then measure the length of the corresponding sides, that is, we can get the length of AB.

Solution: Method: Take a point O that can directly reach point A and point B on the open space, and extend the connecting line AO to D, so that OD = OA will extend the connecting line BO to E, and OE = ob. Connect DE and measure its length, then the length of DE is the distance between A and B, as shown in the figure.

∴△AOB≌△DOE(SAS)？

∴ AB = DE (congruent triangles, corresponding sides are equal).

Example 22 (Henan Curriculum Reform Volume in 2005), as shown in the figure, is a river. Point A is a big tree on the other side, point B is a pole on the shore, and AB is roughly perpendicular to the river bank. At present, the available equipment are: a tape measure and several poles. According to the learned mathematical knowledge, design a scheme to measure the distance between point A and point B, and draw a graph and measurement method on the map.

Hugging: It is difficult to directly measure the distance between A and B, but if you use the method in the above question, this will happen:

The o point is in the middle of the river, which is difficult to get; Even if the O point is selected, the two points of the side CD corresponding to AB in congruent triangles are still on both sides of the river, which are the same as the positions of A and B, so this method is not desirable. To find another way to make the corresponding edge on the shore, just use the method shown below.

Solution: Method: Take two points C and D on the vertical line BE of AB to make CD = BC. Let point D BE the perpendicular DG of BE, and take a point F on DG, so that A, C and F are in a straight line. The measured DF length at this time is the distance between a and B.

Reason: ∵AB⊥BE, DG ⊥ BE ∴∠ B = ∠ BDF = 90.

∴△ABC≌△FDC(ASA)

∴ AB = DF (the corresponding sides of congruent triangles are equal).

Note: Pay attention to distinguish between these two situations, and judge the method according to the specific situation or the language description of the topic. The most obvious difference is that the first case is not vertical, and SAS is used to prove congruence; In the second case, if the triangle is vertical, ASA will be used to prove the congruence of the triangle. Of course, if it is a special case, it needs specific analysis.

As shown in Figure 23, there is a boat A in the river. Set a line segment BC on the shore, and then set two rays BA ′ and CA ′, so that ∠ CBA ′ = ∠ CBA, ∠ BCA ′ = ∠ BCA, so that we can know the length of the distance AB from the ship to point B on the shore by measuring the length of A ′ B. Why?

Tip: If you prove △ BCA ′△ BCA, you will get A ′ B = AB.

Example 24. (Jinhu Experimental Zone, Huai 'an City, 2005)

Known as the figure, Rt△ABC≌Rt△ADE, ∠ ABC = ∠ ade = 900, try to connect two line segments with the point marked with letters as the endpoint. If the two line segments you connect satisfy one of the equal, vertical or parallel relationships, please write and prove it.

Solution: the first one: link CD and BE, and get: CD=BE.

∵△ABC≌△ADE,∴AD=AB,AC=AE

∠CAB =∠EAD；

∴∠cad=∠eab；

∴△ABE≌△ADC .

∴CD=BE。

The second type: connecting DB and CE: DB‖CE,

∵△ABC≌△ADE,∴AD=AB,∠ABC=∠ADE，

∴∠ADB=∠ABD,∴∠BDF=∠FBD

Similarly: ∠FCE=∠FEC,

∴∠FCE=∠DBF，

∴DB‖CE .

The third type: link DB and af; Find af ⊥ b.d,

∵△ABC≌△ADE,∴AD=AB,∠ABC=∠ADE=90 .

And AF=AF, ∴△ADF≌△ABF,

∴∠DAF=∠BAF。

∴AF⊥BD .

The fourth type: connecting CE and AF; Find AF⊥CE,

∵△ABC≌△ADE,∴AD=AB,AC=AE

ABC=∠ADE=90 .

And AF=AF, ∴△ADF≌△ABF,

∴∠DAF=∠BAF ,∴∠CAF=∠EAF .

∴AF⊥BD .

Example 25. (Nanjing, 2005) If point P rotates around fixed point M by180 and coincides with point Q, it is said that point P and point Q are symmetrical about point M, and the fixed point M is called the center of symmetry. At this time, m is the midpoint of the straight line PQ.

As shown in the figure, in the rectangular coordinate system, the coordinates of vertices A, B and O of ⊿ABO are (1, 0), (0, 1) and (0, 0) respectively. Two adjacent points in the point sequence P 1, P2, P3, ... are all symmetric about a vertex of ⊿ABO:

Point P 1 and point P2 are symmetrical about point a, and point P2 and point P3 are symmetrical about point b,

Points P3 and P4 are symmetrical about point O, point P4 and point P5 are symmetrical about point A, and point P5

Point B is symmetrical with point P6, and point O is symmetrical with point P7. symmetrical

The centers are A, B, O, A, B, O, ... These symmetrical centers follow in turn.

The ring. Given that the coordinates of point P 1 are (1, 1), try to find the coordinates of points P2, P7 and P 100.

Prompt: P2 (1,-1) P7 (1, 1) P 100 = (1, -3).

Example 26 (Shenyang, 2005). (1) As shown in Figure 6, how to get Figure B from Figure A by translation or rotation in grid paper, and Figure C from Figure B (translation transformation requires answering the direction and distance of translation; Rotation transformation requires answering the rotation center, rotation direction and rotation angle);

(2) As shown in Figure 6, if the coordinates of points P and P3 are (0,0) and (2 1) respectively, write the coordinates of point P2;

(3) Figure 7 is a part of the designer's design. Please use the method of rotation transformation to rotate the graph clockwise around the O point by 90,180 and 270, and draw the rotated graph in turn. You will get a beautiful pattern, but don't paint it in the wrong place when shading, otherwise it won't have the ideal effect. Come and have a try!

Note: The side length of small squares in grid paper is 1 unit length.