20 18-20 19 school year, dengzhou city, Nanyang city, Henan province, eighth grade (grade one) mid-term mathematics examination paper.
For multiple-choice questions (3 points for each question, ***30 points), please mark the serial number of the only correct answer on the answer sheet.
1. Among the following real numbers, the irrational number is ().
A.π B. C. D
2. The following are true ()
A.= 4 B.= 4 C. = 4 D.=2
3. The following operation is correct ()
a . a 12÷a3 = a4 b .(a3)4 = a 12？
C.(﹣2a2)3=8a5 D.(a﹣2)2=a2﹣4
4. If the product of (x-1) (x2+mx+n) does not contain the quadratic term and linear term of x, then the values of m and n are ().
b.m=﹣2,n= 1 c.m=﹣ 1,n= 1 d . m = 1，n= 1
5. If 2x ~ 3y+z ~ 2 = 0, the value of 16x ~ 82y× 4z is ().
B.﹣ 16
6. An operation is specified: a ※ b = ab+a-b, where a and b are real numbers, then ※ equals ().
A.﹣6 B.﹣2
7. Polynomial ① 4x2-x; ②(x﹣ 1)2﹣4(x﹣ 1)； ③ 1﹣x2； ④ 4x21+4x. After factorization, the result of the same factor is ().
A.① and ② B.③ and ④ C.① and ④ D.② and ③
8. As shown in the figure, △ ABC △ ade, ∠ DAC = 70, ∠ BAE = 100, and BC and DE intersect at point F, then ∠DFB is ().
A. 15
9. As shown in the figure, at △ABC, AD⊥BC, CE⊥AB, the vertical scales are D and E respectively, and AD and CE intersect at point H. Given that Eh = EB = 4 and AE = 6, the length of CH is ().
A. 1
10. As shown in the figure, four congruent rectangles and a small square are combined into a big square. It is known that the area of a large square is 144 and the area of a small square is 4. If a and b represent the length and width of a rectangle respectively (A > B), the following relationship is incorrect ().
a . a+b = 12 b.a﹣b=2 c . ab = 35d a2+B2 = 84
Fill in the blanks (3 points for each small question, *** 15 points)
The square root of 1 1
12. If (a+5) 2+= 0, then a20 18? b20 19=。
13. Calculation: 20132-2014× 2012 =.
14. As shown in the figure, AE⊥AB, AE = AB, BC⊥CD, BC = CD. Please calculate the area of the figure surrounded by solid lines according to the data marked in the figure.
15. Please observe the following formula:
22﹣ 1=3; 32﹣22=5; 42﹣32=7; 52-42 = 9 ... Let n be a positive integer, and use an equation containing n to represent the law you find.
Three. Solve the problem. (* * * 75 points)
16.( 10) Calculate or solve.
( 1)﹣+| 1﹣|﹣(2+)
(2) The arithmetic square root of a number is (2-m-6, and its square root is (2-m), so find this number.
17.(8 points) Decomposition factor.
( 1)4x3y﹣4x2y2+xy3
(2)m3(x﹣2)+m(2﹣x)
18.( 10) (1) calculation: [(ab+1) (ab ﹣ 2) ﹣ (2ab) 2+2] ﹣ (ab).
(2) Simplify first and then evaluate: (x+2) 2+(2x+1) (2x-1)-4x (x+1), where x =-.
19.(9 points) Given that A+B = 3 and AB =-2, find the following values:
( 1)(a﹣ 1)(b﹣ 1)
(2)a2+b2
(3)a﹣b
20.(7 points) As shown in the figure, AB = CD, AE⊥BD and CF⊥BD are known, and the vertical feet are E, F and BF = DE respectively. Verification: AB∨CD.
2 1.( 10) (1) Simplification: (a-b) 2+(b-c) 2+(c-a) 2;
(2) Using the conclusion of (1), and A = 20 15x+20 16, B = 20 15x+20 17, C = 2017.
22.( 10 point) As shown in the figure, it is known that in △ABC, ∠ B = ∠ C, AB = 12 cm, BC = 8 cm, and point D is the midpoint of AB. If point P moves from point B to point C at a speed of 2 cm per second on BC line, at the same time, point Q moves from point B to point C on CA line.
(1) If the moving speeds of point P and point Q are equal, whether △BPD and △CQP are the same after 1 s, please explain the reasons;
(2) If the velocities of point P and point Q are not equal, when the velocity of point Q is what, can △BPD and △CQP be congruent?
23.( 1 1 min) The straight line of CD passing through ∠BCA vertex C, where CA = CB. E and f are two points on a straight CD, and ∠ BEC = ∠ CFA = ∠ α.
(1) If the straight line CD passes through the ∠BCA and E and F are on the ray CD, please solve the following two problems:
① as shown in figure 1, if ∠ BCA = 90 and ∠ α = 90, then be cf (fill in ">", "
② As shown in Figure 2, if 0 < ∠BCA < 180, please add a condition about the relationship between ∠ α and ∠BCA, so that the two conclusions in ① still hold, and prove that the two conclusions hold.
(2) As shown in Figure 3, if the straight line CD passes through the outside of ∠BCA and ∠ α = ∠ BCA, please make a reasonable guess and prove the quantitative relationship among EF, BE and AF.
20 18-20 19 school year, dengzhou city, Nanyang city, Henan province, eighth grade (grade one) mid-term mathematics examination paper.
Reference answers and analysis of test questions
For multiple-choice questions (3 points for each question, ***30 points), please mark the serial number of the only correct answer on the answer sheet.
1. Among the following real numbers, the irrational number is ().
A.π B. C. D
Analysis is based on the definition of irrational numbers.
Solution: a and π are infinite acyclic decimals, that is, irrational numbers;
B, an infinite cyclic decimal, is a rational number;
C = 3, which is a rational number;
D = 4 is a rational number.
So choose: a.
This review mainly examines the definition of irrational numbers, and notes that an infinite number with a radical sign is irrational, and an infinite acyclic decimal number is irrational, such as π, 0.800080008...(65438 adds 0 zeros between every two eights in turn).
2. The following are true ()
A.= 4 B.= 4 C. = 4 D.=2
Analysis can be calculated one by one according to the definitions of arithmetic square root, square root and cube root.
Solution: A. = 4, this option is wrong;
B.= 4, this option is wrong;
C.= 4, this option is correct;
D.≠ 2, = 2, this option is wrong;
So choose: C.
This topic mainly investigates square roots and cubic roots. The key to solving the problem is to master the definitions of square root, arithmetic square root and cubic root.
3. The following operation is correct ()
a . a 12÷a3 = a4 b .(a3)4 = a 12？
C.(﹣2a2)3=8a5 D.(a﹣2)2=a2﹣4
According to the division of the same base power, the sum product of power and the complete square formula, we can analyze them one by one.
Solution: A, A 12 ÷ A3 = A9, this option is wrong;
B, (A3) 4 = A 12, which is correct;
C, (-2a2) 3 =-8a6, this option is wrong;
D, (a ﹣ 2) 2 = a2 ﹣ 4a+4, this option is wrong;
Therefore, choose: B.
This topic mainly examines the operation of algebraic expressions, and the key to solving the problem is to master the formulas of power, power, product and complete square of the same base number.
4. If the product of (x-1) (x2+mx+n) does not contain the quadratic term and linear term of x, then the values of m and n are ().
b.m=﹣2,n= 1 c.m=﹣ 1,n= 1 d . m = 1，n= 1
The polynomial multiplication algorithm is directly used to remove the brackets, and then the equations about m and n are obtained, and then the answers are obtained.
Solution: ∵ (x ﹣1) (product of x2+MX+n) does not contain the quadratic term and linear term of x,
∴(x﹣ 1)(x2+mx+n)
=x3+mx2+nx﹣x2﹣mx﹣n
=x3+(m﹣ 1)x2﹣(m﹣n)x﹣n，
∴,
The solution is m = 1, n = 1,
Therefore, choose: d.
This topic mainly examines the multiplication of polynomials and polynomials, and it is the key to correctly find the coefficients of quadratic terms and linear terms containing X.
5. If 2x ~ 3y+z ~ 2 = 0, the value of 16x ~ 82y× 4z is ().
B.﹣ 16
Find 2x+3y-z according to the problem meaning analysis, and calculate it according to the same base number multiplication and division method.
Solution: ∫2x﹣3y+z﹣2 = 0,
∴2x﹣3y+z=2，
Then the original formula = (24) x \ u (23) 2y× (22) z.
=24x÷26y×22z
=22(2x﹣3y+z)
=24
= 16,
So choose: a.
This topic examines same base powers's division and multiplication operations, and grasps same base powers's division law: the base number is constant, and exponential subtraction is the key to solving problems.
6. An operation is specified: a ※ b = ab+a-b, where a and b are real numbers, then ※ equals ().
A.﹣6 B.﹣2
The analysis can be obtained by calculating = 4, = ﹣ 2, and then calculating according to the operation A ※ B = AB+A ﹣ B specified in the new definition.
※ solution
=4※(﹣2)
=4×(﹣2)+4﹣(﹣2)
=﹣8+4+2
=﹣2,
Therefore, choose: B.
This question examines the mixed operation of real numbers, which belongs to the newly defined question type. The key to solve this problem is to clarify the new definition of the meaning of the problem and the operation order and algorithm of real numbers.
7. Polynomial ① 4x2-x; ②(x﹣ 1)2﹣4(x﹣ 1)； ③ 1﹣x2； ④ 4x21+4x. After factorization, the result of the same factor is ().
A.① and ② B.③ and ④ C.① and ④ D.② and ③
According to the common factor method and the complete square formula, the polynomial of each option is decomposed into factors, and then the results with the same factors can be found.
Solution: ① 4x2-x = x (4x-1);
②(x﹣ 1)2﹣4(x﹣ 1)=(x﹣ 1)(x﹣ 1﹣4)=(x﹣ 1)(x﹣5)；
③ 1﹣x2=( 1﹣x)( 1+x)=﹣(x﹣ 1)(x+ 1)；
④﹣4x2﹣ 1+4x=﹣(4x2﹣4x+ 1)=﹣(2x﹣ 1)2，
② and ③ have the same factors as x → 1,
Therefore, choose: d.
This review mainly examines the factorization of common factors and the factorization of complete square formula. Mastering the formula structure is the key to solve.
8. As shown in the figure, △ ABC △ ade, ∠ DAC = 70, ∠ BAE = 100, and BC and DE intersect at point F, then ∠DFB degree is ().
A. 15
Firstly, we get ∠ b = ∠ d, ∠ BAC = ∠ DAE, so ∠BAD = ∠ CAE, then we get ∠BAD degree, and then the sum of the internal angles of △ABG and △FDG is equal to 180.
Solution: ∫△ABC?△ADE,
∴∠B=∠D,∠BAC=∠DAE，
∠bad =∠BAC∠CAD，∠CAE =∠DAE∠CAD，
∴∠BAD=∠CAE，
∠∠DAC = 70，∠BAE= 100，
∴∠bad=(∠bae﹣∠dac)=( 100 ﹣70)= 15，
At △ABG and △FDG, ∠∠B =∠D, ∠ AGB = ∠ FGD,
∴∠DFB=∠BAD= 15。
So choose: a.
Comments on this issue mainly use the property that congruent triangles's corresponding angles are equal. Pay attention to when solving problems: congruent triangles's corresponding edges are equal, and the corresponding angles are equal.
9. As shown in the figure, at △ABC, AD⊥BC, CE⊥AB, the vertical scales are D and E respectively, and AD and CE intersect at point H. Given that Eh = EB = 4 and AE = 6, the length of CH is ().
A. 1
First, we get ∠ bad = ∠ BCE by using the complementary angle of an equal angle, and then we can prove △ BCE △ hae according to "AAS", so CE = AE = 6, and then we can calculate CE ∠ He.
Solution: ∵AD⊥BC, CE⊥AB,
∴∠BEC=∠ADB=90，
∠∠BAD+∠B = 90，∠BCE+∠B=90，
∴∠BAD=∠BCE，
In △BCE and △HAE.
,
∴△BCE≌△HAE，
∴CE=AE=6，
∴CH=CE﹣HE=6﹣4=2.
Therefore, choose: B.
This topic examines congruent triangles's judgment and nature: congruent triangles's judgment is an important tool to prove that line segments and angles are equal in combination with congruent triangles's nature. When judging the congruence of triangles, the key is to choose appropriate judgment conditions.
10. As shown in the figure, four congruent rectangles and a small square are combined into a big square. It is known that the area of a large square is 144 and the area of a small square is 4. If a and b represent the length and width of a rectangle respectively (A > B), the following relationship is incorrect ().
a . a+b = 12 b.a﹣b=2 c . ab = 35d a2+B2 = 84
Analysis can calculate the side length of a square according to the area of a big square and a small square, then make an equation according to their side length, and make a differential equation according to the fact that the sum of the areas of four rectangles is equal to the areas of two squares.
Solution: A, according to the area of a big square, the side length of the square is 12, then A+B = 12, so option A is correct;
B, according to the area of a small square, it can be found that the side length of the square is 2, then A-B = 2, so option B is correct;
C, according to the sum of the areas of the four rectangles is equal to the area of the big square minus the area of the small square, that is, 4ab = 144-4 = 140 and AB = 35, so option c is correct;
D, (a+b) 2 = A2+B2+2ab = 144, so A2+B2 =144 ﹣ 2× 35 =144 ﹣ 70 = 74, so the D option is wrong.
Therefore, choose: d.
The key to comment on this problem is to correctly analyze the figure and the area formula of the figure, and use the exclusion method to choose.
Fill in the blanks (3 points for each small question, *** 15 points)
The square root of 1 1 It's three.
Analysis can get the answer according to the definition of square root.
Solution: The square root of 8l is 3.
So the answer is: 3.
This topic reviews the knowledge of square root, which belongs to the basic topic, and mastering the definition is the key.
12. If (a+5) 2+= 0, then a20 18? b20 19= 15。
The analysis directly uses the properties of even power and quadratic root to get the values of a and b, and then uses the multiplication algorithm of product to calculate the answer.
Solution: ∫(a+5)2+= 0,
∴a+5=0,5b= 1，
So a =-5, b =,
Then a20 18? b 20 19 =(ab)20 18×b = 1×=。
So the answer is:
This review mainly examines the nature of non-negative numbers and the multiplication operation of products, and mastering the relevant algorithms correctly is the key to solving problems.
13. Calculation: 20132-2014× 2012 =1.
The analysis of 20 14×20 12 is converted into (2013+1)× (2013-1), expanded according to the square difference formula, and then merged.
Solution: Original formula = 20132-(2013+1) × (2013-1)
=20 132﹣20 132+ 12
= 1,
So the answer is: 1.
Note This topic examines the application of the square difference formula. Note: (a+b) (a-b) = a2-b2.
14. As shown in the figure, AE⊥AB, AE = AB, BC⊥CD, BC = CD. Please calculate the area of the figure surrounded by solid lines according to the data marked in the figure.
∠ F = ∠ AGB = ∠ EAB = 90, ∠ FEA = ∠ package. According to AAS certificate △ FEA △ gab, it is deduced that Ag = EF = 6, AF = BG = 2, CG = DH = 2.
Solutions: ∵AE⊥AB, EF⊥AF, BG⊥AG,
∴∠F=∠AGB=∠EAB=90，
∴∠FEA+∠EAF=90，∠EAF+∠BAG=90，
∴∠FEA=∠BAG，
In △FEA and △GAB.
∵,
∴△FEA≌△GAB(AAS)，
∴AG=EF=6,AF=BG=2，
Similarly, CG = DH = 4, BG = CH = 2,
∴FH=2+6+4+2= 14，
∴ The area of trapezoidal EFHD is × (ef+DH )× FH =× (6+4 )×14 = 70.
∴ The shaded area is s- trapezoid EFHD-s △ EFA-s △ ABC-s △ DHC.
=70﹣×6×2﹣×(6+4)×2﹣×4×2
=50.
So the answer is 50.
This topic examines the knowledge points such as the area of triangle, the area of trapezoid, the nature and judgment of congruent triangles, and focuses on transforming the area of irregular figure into the area of regular figure.
15. Please observe the following formula:
22﹣ 1=3; 32﹣22=5; 42﹣32=7; 52-42 = 9 ... Let n be a positive integer, and use an equation containing n to express the law you found (n+ 1) 2-N2 = 2n+ 1.
According to the known equation, the square difference between ordinal number plus 1 and ordinal number is equal to the sum of two sequential numbers plus 1.
Solution: ∫ formula 1 is (1+1) 2:12 = 2×1+1,
The second formula is (2+ 1) 2-22 = 2 × 2+ 1,
The third formula is (3+ 1) 2-32 = 2× 3+ 1,
The fourth formula is (4+ 1) 2-42 = 2× 4+ 1,
The nth formula is (n+ 1) 2-N2 = 2n+ 1,
So the answer is: (n+ 1) 2-N2 = 2n+ 1.
This topic mainly investigates the types of numbers, and the key to solving the problem is to connect the known equation with ordinal number and get the universal law.
Three. Solve the problem. (* * * 75 points)
16.( 10) Calculate or solve.
( 1)﹣+| 1﹣|﹣(2+)
(2) The arithmetic square root of a number is (2-m-6, and its square root is (2-m), so find this number.
Analysis (1) firstly uses the properties of arithmetic square root, cube root and absolute value to get the answer;
(2) Using the definitions of arithmetic square root and square root, find the value of m and get the answer.
Solution: (1) Original formula = 6+3+2- 1-2-2.
=6;
(2) From the meaning of the question: 2m-6 ≥ 0,
∴m≥3,∴m﹣2>0，
So 2m ~ 6 = ~ (2 ~ m),
∴ m = 4, so this number is (2m-6) 2 = 4.
The comment on this question mainly examines the real number operation, and correctly grasping the relevant definitions is the key to solving the problem.
17.(8 points) Decomposition factor.
( 1)4x3y﹣4x2y2+xy3
(2)m3(x﹣2)+m(2﹣x)
When analyzing (1) polynomials with ***3 terms and common factors, the common factors should be extracted first, and then the complete square formula should be considered for decomposition.
(2) Transform the polynomial into m3(x﹣2)﹣m(x﹣2), first extract the common factor, and then consider the square difference formula for decomposition.
Solution: (1) Original formula = xy (4x2-4xy+y2)
=xy(2x﹣y)2
(2) The original formula = m3 (x-2)-m (x-2)
=m(x﹣2)(m2﹣ 1)
=m(x﹣2)(m+ 1)(m﹣ 1)
Comment on this topic, we examine the common factor decomposition method and formula method. Generally speaking, if a polynomial has a common factor, the common factor should be extracted first, and then the formula method should be considered for decomposition.
18.( 10) (1) calculation: [(ab+1) (ab ﹣ 2) ﹣ (2ab) 2+2] ﹣ (ab).
(2) Simplify first and then evaluate: (x+2) 2+(2x+1) (2x-1)-4x (x+1), where x =-.
Analysis (1) first calculates the multiplication in brackets, then merges the similar items, and finally calculates the division;
(2) Calculate multiplication first, then merge similar items, and finally find out by substituting.
Solution: (1) original formula = (A2b2-AB-2-4A2b2+2)-(AB)
=(﹣3a2b2﹣ab)÷(﹣ab)
= 3ab+ 1；
(2) Solution: The original formula = x2+4x+4+4x2- 1-4x2-4x.
=x2+3，
When x =-2, the original formula = (-2) 2+3 = 5.
In this paper, the mixed operation and evaluation of algebraic expressions are investigated, and the correct simplification of algebraic expressions is the key to solve this problem.
19.(9 points) Given that A+B = 3 and AB =-2, find the following values:
( 1)(a﹣ 1)(b﹣ 1)
(2)a2+b2
(3)a﹣b
Analyze (1) expansion formula and substitute it into the whole to get the result;
(2) Using the complete square formula, convert a2+b2 into (a+b) 2-2ab, and substitute the whole into the result;
(3) According to the known result of sum (2), first find the value of (a-b) 2, and then find its square root.
Solution: (1) Original formula = AB-A-B+ 1
=ab﹣(a+b)+ 1
=﹣2﹣3+ 1
=﹣4
(2) original formula = (a+b) 2-2ab
=9+4
= 13
(3)∵(a﹣b)2
=a2+b2﹣2ab
= 13+4
= 17
∴a﹣b=。
This topic reviews the deformation of the whole substitution and the complete square formula. The key to solve this problem is to use the idea of transformation.
20.(7 points) As shown in the figure, AB = CD, AE⊥BD and CF⊥BD are known, and the vertical feet are E, F and BF = DE respectively. Verification: AB∨CD.
Analysis According to congruent triangles's judgment and nature, we can get ∠ b = ∠B=∠D, and according to the judgment of parallel lines, we can get the answer.
The answer proves: ∵AE⊥BD, CF⊥BD,
∴∠AEB=∠CFD=90，
BF = DE，
∴BF+EF=DE+EF，
∴BE=DF.
In Rt△AEB and Rt△CFD,
,
∴Rt△AEB≌Rt△CFD(HL)，
∴∠B=∠D，
∴AB∥CD.
Using the properties of the equation, this paper investigates congruent triangles's judgment and properties, and concludes that be = df is the key to solving the problem, and uses congruent triangles's judgment and properties.
2 1.( 10) (1) Simplification: (a-b) 2+(b-c) 2+(c-a) 2;
(2) Using the conclusion of (1), and A = 20 15x+20 16, B = 20 15x+20 17, C = 2017.
The analysis (1) is simplified according to the mixed algorithm of algebraic expressions and can be substituted for evaluation.
(2) After the original formula is deformed, use the formula of the complete square formula to substitute the known equation into the calculation to get the numerical value.
Solution (1) Solution: Original formula = A2-2ab+B2+B2-2bc+C2-2ac+C2 = 2a2+2b2+2c2-2ac.
(2) solution: the original formula = (2a2+2b2+2c2-2ab-2ac-2bc) = [(a-b) 2+(b-c) 2+(c-a) 2.
When a = 20 15x+20 16, b = 20 15x+20 17, c = 20 15x+20 18,
∴ Original formula = × [(-1) 2+(-1) 2+22] = 3.
This topic reviews the application of factorization, and mastering the complete square formula is the key to solve this topic.
22.( 10 point) As shown in the figure, it is known that in △ABC, ∠ B = ∠ C, AB = 12 cm, BC = 8 cm, and point D is the midpoint of AB. If point P moves from point B to point C at a speed of 2 cm per second on BC line, at the same time, point Q moves from point B to point C on CA line.
(1) If the moving speeds of point P and point Q are equal, whether △BPD and △CQP are the same after 1 s, please explain the reasons;
(2) If the velocities of point P and point Q are not equal, when the velocity of point Q is what, can △BPD and △CQP be congruent?
Analysis (1) According to the equal velocity of P and Q, after 1 s, the congruences of △BPD and △CQP can be obtained by SAS;
(2) According to BP≠CQ, △BPD △CQP, BP = CP = 4, and then T = 2, A = 3, that is, when the speed of point Q is 3cm/s, △ BPD and △ CQP can be congruent.
Solution: (1)△BPD and △CQP are congruent.
Reason: ∫t = 1 sec,
∴BP=CQ=2，
∴CP=8﹣BP=6，
∫AB = 12，
∴BD= 12×=6，
∴BD=CP，
∠ b =∠ c，
∴△bpd≌△cqp(sas)；
(2)∵BP≠CQ，△BPD?△CQP，
∴BP=CP=4，
∴t=2，
∴BD=CQ=at=2a=6，
∴a=3，
∴ When the velocity of Q point is 3cm/s, △BPD and △CQP can be congruent.
This topic reviews the nature and judgment of congruent triangles, and the key to solve this problem is to find △ BPD △ CQP by solving a linear equation of one variable. Note: congruent triangles has equal sides and equal angles.
23.( 1 1 min) The straight line of CD passing through ∠BCA vertex C, where CA = CB. E and f are two points on a straight CD, and ∠ BEC = ∠ CFA = ∠ α.
(1) If the straight line CD passes through the ∠BCA and E and F are on the ray CD, please solve the following two problems:
(1) as shown in figure 1, if ∠ BCA = 90 and ∠ α = 90, then be = CF (fill in ">", "
② As shown in Figure 2, if 0 < ∠BCA < 180, please add a condition about the relationship between ∠ α and ∠ BCA ∠ α+∠ ACB = 180. , so that the two conclusions in ① are still valid and proved.
(2) As shown in Figure 3, if the straight line CD passes through the outside of ∠BCA and ∠ α = ∠ BCA, please make a reasonable guess and prove the quantitative relationship among EF, BE and AF.
Analysis (1)① Find out ∠ BE=CF = ∠ AFC = 90, ∠ CBE = ∠ ACF, and deduce BE = CF and CE = AF according to AAS certificate.
② Find out ∠ BE=CF = ∠ AFC, ∠ CBE = ∠ ACF, and deduce BE=CF, BE = CF and CE = AF according to AAS proof.
(2) Find out ∠ BE=CF = ∠ AFC, ∠ CBE = ∠ ACF, and deduce BE=CF, BE = CF and CE = AF according to AAS certificate.
Solution: (1)① As shown in figure 1,
Point e is to the left of point f,
∵BE⊥CD,AF⊥CD,∠ACB=90，
∴∠BEC=∠AFC=90，
∴∠BCE+∠ACF=90，∠CBE+∠BCE=90，
∴∠CBE=∠ACF，
At △BCE and △CAF,
,
∴△BCE≌△CAF(AAS)，
∴BE=CF,CE=AF，
∴EF=CF﹣CE=BE﹣AF，
When e is on the right of f, it can be proved that ef = af-be,
∴ef=|be﹣af|；
So the answer is =, ef = | be-af |.
② When ∠ α+∠ ACB = 180, the two conclusions in ① still hold;
Proof: As shown in Figure 2,
∠∠BEC =∠CFA =∠a，∠α+∠ACB= 180
∴∠CBE=∠ACF，
At △BCE and △CAF,
,
∴△BCE≌△CAF(AAS)，
∴BE=CF,CE=AF，
∴EF=CF﹣CE=BE﹣AF，
When e is on the right of f, it can be proved that ef = af-be,
∴ef=|be﹣af|；
Therefore, the answer is ∠ α+∠ ACB = 180.
(2) Conclusion: EF = Be+AF.
Reason: As shown in Figure 3,
∠∠BEC =∠CFA =∠a∠a =∠BCA，
And ∵∠ EBC+∠ BCE+∠ BEC = 180, ∠ BCE+∠ ACF+∠ ACB = 180,
∴∠EBC+∠BCE=∠BCE+∠ACF，
∴∠EBC=∠ACF，
At △BEC and △CFA,
,
∴△BEC≌△CFA(AAS)，
∴AF=CE,BE=CF，
∫EF = CE+CF，
∴EF=BE+AF.
This review comprehensively examines the knowledge of triangle synthesis, congruent triangles's judgment and nature. The key to solving the problem is to master congruent triangles's judgment and nature, and pay attention to the graphic changes of such problems. The conclusions are basically the same, and the proof methods are completely similar, which belongs to the common questions in the senior high school entrance examination.