First, multiple-choice questions: (8 questions in this big question, out of 24).
1.c； 2.b； 3.c； 4.a； 5.b； 6.d；
Two. Fill in the blanks: (This big question is *** 12, out of 48 points)
7.-5; 8.； 9. Decline; 10.； 1 1.∠ B = ∠ ACD, etc. 12.； 13.30; 14.； 15.； 16. 150; 17.8; 18.3:4;
Three. (There are ***7 questions in this big question, out of 78 points)
19. (The full mark of this question is 10) Solve the equation:
Solution: scores. Or ... -Subtract 2 points from (2).
The original equations can be simplified to-2 points.
By solving these two equations, the solution of the original equation is-6 points.
(Problems solved by substitution elimination method are graded step by step accordingly.)
20. (Full score of this question 10) Proof: ∵BA? BD=BC？ Become ∴-1.
∫BD shares ∠ ABC ∴∠ Abe = ∠ CBD-.
∴△ Abbe ∽△ CBD-.
∴∞-e =∠BDC。
∠∠ade =∠BDC∴∠e =∠ade-。
∴。
2 1. (Full score for this question 10, 5 points for each small question)
(1) proves that ∵AB is the diameter and CD ⊥ AB ∴ CF = DF-.
∴ PC = PD -。
(2) coupling OE,-.
Pe = oe = oc, ∠ APC = 20 ∴∠ EOP = ∠ APC = 20, ∠ OCP = ∠ OEC = 40-2 points.
∴∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠873
22. (The full mark of this question is 10) Solution: If you pass point C, you will get CF//AB; if you pass point F, you will get FG⊥AB and -2 points at point G.
∫DE is perpendicular to CD,
∴ In the right triangle DFC, 1 min.
cd = 3m，∴ CF = 6m -。
According to the meaning of the question, the quadrilateral FCBG is a rectangle ∴ CF = BG = 6m, BC=FG.
AB = 28m, e is the midpoint of AB, ∴ EG = 14-6 = 8m-2 points.
In the right triangle EFG,
∴∴∴ FG = m -2 points.
BC = m
Answer: When the pole is high, the best lighting effect can be achieved. - 1.
23. (The full score of this question is 12, and each small question is 6 points. )
(1) similar triangles: △AEM∽△BMG, △FEM∽△FMA,-2 points.
The following proof △AEM∽△BMG
∵ RT-△ ABC，∠ ACB = 90，AC = BC，∴∠ A = ∠ B = 45 - 1。
∠∠EMB =∠EMG+∠GMB =∠A+∠AEM∠EMG = 45
∴∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠ - .
∴△ AMF ∽△ BGM。 - .
(2) In Rt△ABC, ∠ ACB = 90, AC = BC = 4.
∴ AB = -。
∫m is the midpoint of AB, ∴ am = BM = minutes.
∫△AMF∽△bgm,∴
∴ - .
∴, - .
∴ - .
24. (Full score for this question 12, 4 points for each small question)
Solution: (1) Intersection point A represents AH⊥BC in h-.
∫A's coordinates are (2,2), AB=AC, BC=8,
∴BH=CH=4, ∴ B (0 0,6), C (0 0,2)-2 points.
∵AH//OD,∴
∴ d (3，0) -。
(2) The parabola passes through points A (2 2,2), C (0 0,2) and D (3 3,0).
According to the meaning of the question, you can get: solution:-3 points.
Therefore, the score of sub-resolution function is-1.
(3) at point c, CE⊥AB is at e-.
∵
Ab =, BC=8, AH = 2 ∴-2 points.
In right triangle CAE, ∠ CAD =-1.
(3 points for obtaining CE by other methods)
25. (The full mark of this question is 14, item (1) is 4, item (2) is 5, and item (3) is 5).
Solution: (1) Take the midpoint g of BC, followed by AG,-.
∵ The circumscribed circle of circle A and circle G, ∴ Ag = AE+ 1-.
∫ In square ABCD, AB=2, AE=
In the right triangle ABG, it is 65438+.
∴ (negative numbers are discarded)-1 minute.
∴ When a circle with A as the center and AE as the radius circumscribes a circle with BC as the diameter, the length of AE is.
(2) Point D is DH⊥PE in H, and DF is even-.
∫PD = PE ∴∠pde=∠ped
∫ square ABCD ∴DC//AB ∴∠PDE=∠DEA
∴∠ped=∠dea≈a =∠dhe = 90，DE=DE
∴△ DAE △ dhe ∴ da = DHEA = eh-1min.
dc = dh，∠DCF =∠DHF = 90° df = df ∴△dhf≌△dcf
∴ CF = FH - 65438。
ae =，CF= ,∴
In the right triangle BEF, Ⅷ
After the end, we got:-.
(3)∫ef =，∴ ∴
Solution:-.
When,
∵B falls on the plane b' and there is a vertical foot q along the straight line EF, and ∴∴∴ bb' ⊥ ef.
∴BQ=，B B'=
∵E and Q are the midpoint of AB and BB' respectively, ∴ EQ//AB' ∴ A BB' = ∠ EQB = 90.
In △AB'B and △BEF,
∴ = ∴ △ AB 'b ∽ △ BEF -。
(δAB' b∽△BEF can also be proved by similarity transitivity, or graded step by step. )
When,
∵ eq//ab' ∴△ a bb' is not a right triangle.
∴△AB'B is not like△ △BEF-.
To sum up, when EF=, △AB'B∽△BEF
△AB'B is not similar to △ △BEF when EF= 0.