1. The following judgment is correct ()

Two triangles with two sides and a diagonal line are congruent.

B there are two isosceles triangles with equal sides and an included angle of 30 degrees.

C Two right-angled triangles with one angle and one side are congruent.

D the opposite side with two angles and one angle corresponds to the congruence of two triangles.

2. As shown in figure 1, △ABC and △BDE are equilateral triangles, AB.

A.ae = CD B.AE & GTCD C.AE & LTCD D. Not sure.

3. As shown in Figure 2, in equilateral △ABC, d, e and f are points (not midpoints) on AB, BC and CA respectively, and AD=BE=CF, and the number of congruent triangles groups in the figure is ().

A3 Group B.4 Group C.5 Group D.6 Group

4. As shown in Figure 3, D is the midpoint of AB side of △ABC, the intersection D is DE‖BC, AC to E, and the point F is on BC, so △DEF and △DEA are congruent. The number of such points f is ().

A.4 B.3 C.2 D. 1

5. The following proposition is wrong ()

A. the rectangle is a parallelogram; Similar triangles must be congruent triangles.

C. The diagonal lines of the isosceles trapezoid are equal. D. Two straight lines are parallel and the isosceles angles are equal.

6. The following proposition, the correct proposition is ()

A. A quadrilateral with equal diagonals is a rectangle; Two isosceles triangles with equal base angles are congruent.

A diagonal line divides a parallelogram into two similar triangles.

D. The circle is a centrally symmetric figure, not an axisymmetric figure.

7. The following proposition belongs to the false proposition is ()

A. isosceles isosceles triangle; The two base angles of an isosceles triangle are equal.

C, the midline on the bottom edge of the isosceles triangle coincides with the height on the bottom edge; An isosceles triangle is a figure with central symmetry.

8. Among the following true propositions, the inverse proposition is true ().

A. The corresponding angles of congruent triangles are equal.

B. If two figures are symmetrical to each other, they are conformal.

C. an equilateral triangle is an acute triangle.

In a right triangle, if an acute angle is equal to 30, then the right side it faces is equal to half of the hypotenuse.

9. As shown in Figure 4, it is known that in △ABC, AQ=PQ,PR=PS,PR⊥AB is in R,PS⊥AC is in S, so there are three conclusions: ① As = AR; ②QP AR； ③△BRP?△QSP()

A. All correct B. Only ① and ② are correct; C. Only ① is correct D. Only ① and ③ are correct.

10. Observe the picture below and read the relevant words below, as shown:

Two straight lines intersect, three straight lines intersect, four straight lines intersect, with at most one intersection point and at most three intersections; Up to 6 intersections, like this, 10 straight lines intersect, and the maximum number of intersections is ().

40 dollars, 45 cents, 50 cents, 55 cents

1 1. The condition for making two right triangles congruent is ()

A. acute angle corresponds to equation b, and edge corresponds to equation B.

C. the two acute angles are equal. D. Two right-angled sides are equal.

12. Among the following conditions, the condition that two triangles cannot be congruent is ()

A. the angles on both sides are equal; B. Two corners and one edge are equal.

C. Trilateral equality; D, two sides and their included angles are equal.

Two. Fill in the blanks: (16, 3 points, the rest 1 point, 40 points * * *)

13. As shown in Figure 6, △ OCA △ OBD, ∠C and ∠B, ∠A and ∠D are corresponding angles, then another set of corresponding angles is _ _ _ _ and _ _ _ _ _ _, and the corresponding edges are _ _.

14. In △ABC and △KMN, AB=KM, AC=KM, ∠A=∠K, then △ ABC _ _, ∠ C = _ _ _ _.

15. As shown in Figure 7, △ABC≌△EFC,BC=FC,AC⊥BE, then AB = _ _ _ _, AC = _ _ _, ∠ B = _ _ _ _ ∠ A = _ _ _ _

16. As shown in Figure 8, AD⊥BC,DE⊥AB,DF⊥AC,D, E and F are vertical feet, and BD=CD, then congruent triangles in the figure has _ _ _ _ _ _ _ _ _ _.

17. As shown in Figure 9, △ ABC △ ade, ∠ B = 30, ∠ EAD = 24, ∠ C = 32, ∠ D = _ _ _ ∠ DAC = _.

18. In △ABC, ∠ A = 90, CD is the bisector of ∠C, AB is at point D, and DA=7, then the distance from point D to BC is _ _ _ _ _.

19. The proposition that "two lines perpendicular to the same line are parallel" is _ _ _ _ _ _ _ _ _ _ _ _ _ _ _.

20. Proposition: The conclusion that "two lines parallel to the same line are equal" is _ _ _ _ _ _ _ _ _ _ _ _.

2 1. Write the proposition "the complementary angles of equal angles are equal" as "if"

22. As shown in figure 10, the reason that should be paid attention to after "this figure ∠ 1=∠4, so BD‖AC" is _ _ _ _ _ _.

23. As shown in figure 1 1, AB=DC is known. According to the congruence identification method (SAS), in order to make △ ABC △ DCB, only one condition _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _.

24. as shown in figure 12, when ⊙O and ∠ BOC = 70, rotate △AOC clockwise by _ _ _ _ degrees to coincide with △ _ _ _ _, so △ _ _ _ ⊙; △_ _

25. as shown in figure 13, line segments AC and BD intersect at point o, OA = OC, AE ‖ FC, BE = FD, then there are _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ in the figure.

26. Fold a 20cm long metal wire into a triangle whose three sides are integers, so the number of unequal triangles is _ _ _ _ _ _ _.

27. as shown in figure 14, rotate △ABC counterclockwise around point a to get △ADE, then AB = _ _ _ _ _, BC = _ _ _ _ _, AC = _ _ _ _, ∠ B = _ _ _ _ ∠ C =

28. as shown in figure 15, in △ABC and △ABD, ∠ c = ∠ d = 90, and in order to make △ ABC △ Abd, a condition needs to be added: _ _ _ _ _ _ _.

29. as shown in figure 16, AB = DC, AD = BC, ∠ 1 = 50, ∠ 2 = 48, then the degree of ∠B is _ _ _ _.

Three. Problem solving: (6 points for each question, ***36 points)

30. Judge whether the following propositions are true or false. If it is false, please give a counterexample.

(1) An isosceles triangle with an angle of 60 is an equilateral triangle.

(2) A triangle with two acute angles is an acute triangle.

3 1. As shown in the figure, it is known that CD⊥AB is at point D,BE⊥AC is at point E, BE and CD intersect at point O, and AO bisects ∠BAC.

Verification: OB=OC.

32. As shown in the figure, it is known that points A, E, F and D are on the same straight line AE=DF,BF⊥AD,CE⊥AD, and their footsteps are F, E and BF = CE respectively. Verification: AB‖CD.

33. As shown in the figure, it is known that ∠DBC=∠ACB, ∠ Apo =∠DCO, and verification: AO=DO.

34. As shown in the figure, it is known that in quadrilateral ABCD, e is a point on AC, ∠BAC=∠DAC, ∠BCA= ∠DCA.

Verification: ∠DEC=∠BEC.

35. As shown in the figure, AB=AE, ∠ABC=∠AED, BC=ED, and point F is the midpoint of the CD.

(1) verification: af ⊥ CD;

(2) What new conclusions can you draw after connecting BE? Please write three (no proof is needed).

Four, comprehensive questions within the discipline: (6 points)

36. As shown in the figure, it is known that AB is the diameter ⊙O, C and D are two points on the circle, CE⊥AB,DF⊥AB, and the vertical foot is E and F respectively, which proves that CE=DF.

Verb (abbreviation of verb) expands inquiry: ((1) 2 points, (2) 6 points, ***8 points)

37. As shown in the figure, the two ends of the intersection line AB are straight lines L 1‖L2, and the bisector of the inner angle on the same side intersects at point E, and the intersection point E is straight lines DC that intersect with straight lines L 1 and L2 respectively, and points D and C are on the same side of AB and do not overlap with A and B. 。

(1) Measure and compare whether AD+BC and AB are equal with compasses and straightedge, and write your conclusion;

(2) Analyze the conclusion with the learned principles and reveal the law.

Six, interdisciplinary comprehensive questions: (6 points)

38. As shown in the figure, it is known that when the distance between the object A'b' and the focal length of the convex lens is twice, that is, AO=2f, the image A'b' is measured in an anti-isometric way. Find the relationship between image distance OA' and F.

Answer:

One,

1.D

Guidance: This question examines the identification of congruence of two triangles, and the training should be strengthened.

2.A

Solution: ∫△ABC and △BDE are equilateral triangles, ∴ △ DBE = ∠ ABC = 60, AB = BC, BE = BD.

∴∠DBE+∠CBE=∠ABC+∠CBE, that is ∠CBD=∠ABE,

In △ABE and △CBD, AB=CB, ∠ABE=∠CBD, BE=BD,

∴△ABE≌△CBD,∴AE=CD.

Pointing: It is a common method to prove that two straight lines are equal with two triangles, which students should master.

3.C

Solution: The congruent triangle in the graph is: △ adg △ beh △ cfn; △ABH?△BCN?△CAG； △ABE?△BCF?△CAD； △ABF?△CAE?△BCD； △AHF?△BND?△CGF； * * * There are 5 groups.

Hugging: Finding congruent triangles correctly according to the topic is the key point of this topic, and students tend to miss some congruent triangles.

4.D

Solution: As shown in the answer sheet, in order to make △ DEF △ DEA, it is necessary to make DF‖AC pass through point F BC or EF'‖AB pass through point F BC through point E. Inferred from the triangle midline theorem, both f and F' are the midpoint of BC, so the two points coincide.

Hugging: This problem is an application of the derivation of the triangle median theorem.

5.B

Clue: congruence between two triangles is a special case of similarity between two triangles, so congruence must be similar, but similarity is not necessarily congruence.

6.c solution: in ABCD, ABCD, BC‖AD,

∴△abd∽△cdb. ∴∠adb=∠dbc,∠abd=∠cdb

Hug: A diagonal of the parallelogram divides the parallelogram into two triangles, which are not only similar, but also congruent.

7.D

Hugging: Because the isosceles triangle is "three lines in one", students can easily mistake it for a central symmetrical figure.

8.D solution: As shown in the answer sheet, at Rt△ABC, ∠ ACB = 90, BC = AB, take the midpoint D of AB and connect the CD.

∴CD=DB= AB,∴CB=CD=BD, that is, △BCD is an equilateral triangle.

∴∠B=60 ,∴∠A=90 -∠B=90 -60 =30。

Hugging: Correctly distinguishing the topic and conclusion of the original proposition is the key to writing its inverse proposition.

9.B

Solution: As shown in the answer sheet, ∵PR⊥AB,PS⊥AC,∴△APR and △APS are right triangles.

In Rt△APR and Rt△APS, PR = PS and AP=AP,

∴rt△apr≌rt△aps,∴ar=as,∠par=∠ pass,

∵aq=pq,∴∠pas=∠apq,∴∠par=∠apq,∴qp‖ar.

Hugging: This problem is an application of the theorem of congruence and isometric line of two triangles in geometry.

10.B

Solution: The fourth straight line intersects with the first three straight lines at most, plus three intersections, the fifth straight line intersects with the first four straight lines at most, plus four intersections ... The tenth straight line intersects with the first nine straight lines at most, plus nine intersections, so the number of intersections with 10 straight lines is: 1+2+3+...+9 = 45.

Guidance: with the increase of the number of straight lines, the maximum intersection points also increase; Every time a straight line is added, the maximum number of intersections is the same as the original number of straight lines. Pay attention to observation and summary.

1 1.D

12. One instruction: When applying the identification method of two triangles, students easily mistake (SSA) for a judgment method.

Second,

13.AOC and ∠ DOB; OA and od; OC and ob; Ac and decibel

14.△KMN； ∠N。

15.EF； EEC; Treaty on Conventional Armed Forces in Europe; CEF。

16.△ABD?△ACD，△ADE?△ADF，△BDE?△CDF

17.36 ; 24

(13 ~ 17) Guidance: When answering congruent triangles's question, we must correctly use its identification method and characteristics to solve it, and master the method of finding the corresponding edges and angles skillfully.

18.7 direction: it can be obtained from the properties of the angular bisector.

19. Two straight lines are perpendicular to the same straight line.

20. Two straight lines are parallel.

2 1. If two angles are equal, their complementary angles are also equal.

(Question 19 ~ 2 1) Hint: These three questions are an investigation of the composition of the proposition, and students should be guided to distinguish the conclusion of the proposition from the topic setting and use it correctly.

22. The internal dislocation angles are equal and the two straight lines are parallel. Pointing: It is very important for beginners to mark the reasons when proving, which is conducive to getting familiar with the theorem and deepening the understanding and application of the theorem.

23.∠ABC=∠DCB

24.70 ; BODAOC bode

25.3; △AOE?△COF、△AOB?△COD、△CDF?△ABE。

Question (23-25): The above questions are all applications of congruence of two triangles. Note that when two triangles are congruent, the opposite sides of equal angles must be corresponding sides.

26.8 Guidance: This question is actually to find the number of triangles with a circumference of 20cm from the data of 1cm, 2cm, 3cm, 4cm, 5cm, 6cm, 7cm, 8cm and 9cm.

27.AD； De; AE； ∠D； ∠E； ∠DAE。

28.BC=BD (just fill in one condition that meets the requirements)

29.82 (Questions 27 ~ 29): The above questions are also the application of two triangle congruence questions, and students are prone to make mistakes when looking for the corresponding angles and sides.

Third,

30.( 1) true proposition; (2) false proposition. For example, in △ABC, ∠ A = 20, ∠ B = 30, ∠ C = 130, then △ABC is an obtuse triangle.

Hugging: It is very important to understand the proposition correctly and judge whether it is true or not.

3 1. Proof: As shown in the answer sheet: ∵CD⊥AB,BE⊥AC,∴∠ODA=∠OEA.

* OA stock ∠∴∠bao=∠cao BAC,

And OA=OA,∴△OAD≌△OAE,∴OD=OE,

In △OBD and △OCE, OD=OE, ∠ODB=∠OEC, ∠BOD=∠COE,

∴△OBD≌△OCE,∴OB=OC.

Tip: This problem is solved by two congruences, and readers often use △ OAB △ OAC to solve it directly.

32. Proof: ∫∠DBC =∠ACB, ∠ ABO =∠DCO,

∴∠DBC+∠ABO=∠ACB+∠DCO, that is ∠ABC=∠DCB,

And ∠ ACB = ∠ DBC, BC = CB, ∴△ACB?△DBC, ∴ AB = DC.

* ABO =∠dco，∠AOB=∠DOC,∴△ABO≌△DCO,∴OA=OD.

Hugging: This question is proved by two identical questions, and students can easily mistake it for △ ABO △ CDO.

33.

34. Proof: in △ABC and △ADC, ∠BAC=∠DAC, AC=AC, ∠BCA=∠DCA.

∴△BAC≌△DAC,∴BC=DC.

In △DCE and △BCE, EC=EC, ∠DCE=∠BCE, CD=CB,

∴△DCE≌△BCE,∴∠DEC=∠BEC.

Hugging: We should carefully observe the figure, correctly find out the proved congruent triangles from the figure, and flexibly choose and use the method of judging the congruence of two triangles.

35.( 1) Proof: As shown in the answer sheet, connect AC, AD,

In △ABC and △AED, AB=AE, ∠ABC= ∠AED, BC=ED,

∴△ABC≌△AED,∴AC=AD，

∵FC=FD,∴AF⊥CD.

(2) Be ⊥ AF, Be ‖ CD and △ Abe are isosceles triangles.

Hugging: This question is a comprehensive application of proving and exploring questions in geometry, which is helpful to cultivate our awareness of exploring.

Fourth,

36. Proof: ∫,∴ac=bd.

∵CE⊥AB,DF⊥AB,∴∠CEA=∠DFB=90，

∫ab is the diameter, ∴, ∴∠ A = ∠ B.

In △AEC and △BFD, AC = BD, ∠ CEA = ∠ DFB = 90, ∠ A = ∠ B.

∴△AEC≌△BFD,∴EC=FD.

Hugging: This topic is a comprehensive application of two triangles in a circle, which further strengthens the connection of knowledge in the topic.

Five,

37.( 1) Solution: AD+BC=AB

(2) As shown in the answer sheet, extend the intersection of AE and point F,

∫l 1‖L2 ,∴∠ 1=∠f，

∫≈ 1 =∠2, ∴∠ 2 =∠ F, ∴ BA = BF, ∴△ BAF is an isosceles triangle.

∵∠3=∠4,∴EA=EF.

In △AED and △FEC, ∠ 1= ∠F, AE=FE, ∠5=∠6,

∴△AED≌△FEC,∴AD=CF.

∵BF=BC+CF,∴BF=BC+ AD, so BC+AD=AB.

Hugging: This question is a comprehensive inquiry question in geometry, which should be carefully analyzed to strengthen the communication and connection between knowledge points.

Six,

38. solution: in △AOB and △ a ′ ob ′,

∫AB = A′B′，∠BAO =∠B′A′O，∠BOA =∠B′OA′，

∴△aob≌△a′ob′,∴oa′=oa.

∵oa=2f,∴oa′=2f.