Dynamics is a hot issue in the senior high school entrance examination in recent years, including point motion, linear motion and plane motion. To solve this kind of problem, we should "use static braking", that is, turn the dynamic problem into a static problem, which is a special case of the dynamic problem. Common types of questions include maximum problem, area problem, sum and difference problem, fixed value problem and existence problem. We have discussed the maximum problem, the area problem and the sum difference problem, and this topic is about the fixed value problem.
First, the line segment (sum and difference) is a fixed value problem:
Typical example:
Example 1: (20 12 Suihua 8 o'clock in Heilongjiang) As shown in the figure, point E is a point on the diagonal BD of rectangular ABCD, BE=BC, AB=3, BC=4, point P is a point on the straight EC, PQ⊥BC is at point Q, and PR⊥BD is at point R.
(1) As shown in figure 1, when point P is the midpoint of line segment EC, it is easy to prove: PR+PQ= (without proof).
(2) As shown in Figure 2, when point P is any point on the line segment EC (not coincident with points E and C) and other conditions remain unchanged, does the conclusion in (1) still hold? If yes, please give proof; If not, please explain why.
(3) As shown in Figure 3, when point P is any point on the extension line of line segment EC, other conditions remain unchanged, what is the quantitative relationship between PR and PQ? Please write down your guess directly.
Answer: (2) The conclusion PR+PQ = in Figure 2 still holds. Proved as follows:
Connect BP so that CK⊥BD is at point K after point C.
∵ quadrilateral ABCD is a rectangle, ∴∠ BCD = 90.
CD = AB = 3，BC=4，∴.
∫S△BCD = BC？ CD=BD？ CK,∴3×4=5CK,∴CK=。
∫S△BCE = BE？ CK，S△BEP=PR？ BE，S△BCP=PQ？ BC, and s △ BCE = s △ BEP+s △ BCP,
∴BE？ CK = public relations? BE+PQ？ In 200 BC.
∵BE=BC，∴ CK = PR+PQ。 ∴CK=PR+PQ。
ck =，∴ PR+PQ =。
(3) The conclusion in Figure 3 is that PR-PQ =.
Example 2: (20 12 Jiangxi Province 10) As shown in the figure, it is known that the quadratic function L 1: y = x2-4x+3 intersects the X axis at two points A and B (point A is on the left of point B) and the Y axis at point C. 。
(1) Write the opening direction, symmetry axis and vertex coordinates of the quadratic function L 1;
(2) Study the quadratic function L2: y = kx2-4kx+3k (k ≠ 0).
(1) Write two identical properties of images related to quadratic function L2 and quadratic function L 1;
② Is there a real number k that makes △ABP an equilateral triangle? If it exists, request the value of k; If it does not exist, please explain the reason;
③ If the straight line y=8k intersects with parabola L2 at points E and F, does the length of line segment EF change? If not, request the length of EF; If yes, please explain why.
Solution: (1)∵ parabola,
∴ The opening of the quadratic function L 1 is upward, the symmetry axis is a straight line x=2, and the vertex coordinate is (2, ∴ 1).
(2)① Quadratic functions L2 and L 1 have two identical properties about images:
The symmetry axis is x = 2；; Both pass through a (1, 0) and b (3, 0).
② There is a real number k, which makes △ABP an equilateral triangle.
Vertex P(2, -k).
∵A( 1,0),B(3,0),∴AB=2
To make △ABP an equilateral triangle, |-k |-k|=,
∴k= .
③ The length of line segment EF will not change.
The straight line y=8k intersects the parabola L2 at two points e and f,
∴kx2﹣4kx+3k=8k,∵k≠0,∴x2﹣4x+3=8。 Solution: x 1 =- 1, x2=5.
∴EF=x2﹣x 1=6。 ∴ The length of line segment EF will not change.
Example 3: (20 12, Dezhou, Shandong, 12) As shown in the figure, there is a square piece of paper ABCD with a side length of 4, and the point P is a point on the side of the square AD (not coincident with points A and D). Fold the square paper so that point B falls on point P, point C falls on point G, and PG passes through DC at point H, and the crease is EF, connecting BP and D.
(1) Verification: ∠ APB = ∠ BPH;
(2) When the point P moves on the edge AD, does the perimeter of △PDH change? And prove your conclusion;
(3) Let AP be X and the area of quadrilateral EFGP be S, and find out the functional relationship between S and X. Is there a minimum value for S? If it exists, find this minimum value; If it does not exist, please explain why.
Solution: (1) as shown in figure 1, pe = be, ∴∠ EBP = ∠ EPB.
∠∠Eph =∠EBC = 90，
∴∠Eph∠EPB =∠ebc∠ebp, that is ∠PBC=∠BPH.
And BC ∫ AD ∨, ∴ APB = ∠ PBC. ∴∠ APB = ∠ BPH.
(2) The circumference of △ PHD does not become a fixed value of 8. Proved as follows:
As shown in Figure 2, B is BQ⊥PH, and the vertical foot is Q.
According to (1), ∠APB=∠BPH,
∠∠A =∠BQP = 90，BP=BP，
∴△ABP≌△QBP(AAS)。 ∴AP=QP,AB=BQ。
ab = bc，∴BC=BQ.
∫∠c =∠bqh = 90，BH=BH，
∴△BCH≌△BQH(HL)。 ∴CH=QH。
The circumference of ∴△PHD is: PD+DH+PH=AP+PD+DH+HC=AD+CD=8.
(3) As shown in Figure 3, if F is FM⊥AB and the vertical foot is M, then FM=BC=AB.
And ∵EF is a crease, ∴EF⊥BP.
∴∠EFM+∠MEF=∠ABP+∠BEF=90 .∴∠EFM=∠ABP。
∠∠a =∠emf = 90，AB=ME，∴△EFM≌△BPA(ASA).
∴EM=AP=x.
∴, (4-be) 2+x2 = be2 in Rt△APE, that is.
∴。
Also, quadrilateral PEFG and quadrilateral BEFC are congruent,
∴。
When x=2, the minimum value of s is 6.
Example 4: (20 12, Quanzhou, Fujian, 12) It is known that A, B and C are not in a straight line.
(1) If points A, B and C are all on ⊙O with radius r,
I) as shown in figure 1, when ∠ A = 45 and R= 1, find the number of times of ∠BOC and the length of BC;
Ii) As shown in Figure 2, when ∠A is an acute angle, it is proved that sin∠A =；;
(2). If the two endpoints of the fixed line BC slide at AM and AN on both sides of ∠MAN respectively (B and C do not coincide with point A), as shown in Figure 3, when ∠ Man = 60, BC=2, BP⊥AM, CP⊥AN, and the intersection point is point P, try. Please explain the reason.
Answer: (1)I)∫∠A = 45,
∴∠BOC = 90° (the circumferential angle of the same arc is equal to half the central angle of the same arc).
And ∵R= 1, ∴ BC= comes from Pythagorean theorem.
Ii) proof: connect BO and extend, intersect at point e, and connect EC.
It is known that EC⊥BC (the circumferential angle of the diameter is 90),
And ∠E=∠A (the circumferential angles of the same arc are equal).
So sin∠A=sin∠A=
(2) remain unchanged. The reason for this is the following：
As shown in the figure, connect AP, take the midpoint k of AP, and connect BK and CK.
In Rt△APC, CK=AP=AK=PK.
Similarly: BK=AK=PK.
∴CK=BK=AK=PK。 Points a, b, p and c are all on K.
From (1)ii)sin∠A= we can know that sin60 =.
∴AP= (fixed value).
Example 5: A(-20 12 Weifang, Shandong 1 1) As shown in the figure, it is known that the parabola and the coordinate axis intersect at three points A(-2, o), B(2, 0) and C(0, -L) respectively, and the straight line passing through the coordinate origin O is y=kx.
(2) Verify that the circle with the diameter of ON is tangent to the straight line;
(3) Find the length of line segment MN (represented by k) and prove that the sum of the distances from m and n points to a straight line is equal to the length of line segment MN.
Solution: (1) Let the quadratic resolution function corresponding to the parabola be y = AX2+BX+C.
A solution was obtained.
Parabola corresponds to the analytical expression of quadratic function, so.
(2) Let M(x 1, y 1) and N(x2, y2), because points m and n are on a parabola,
∴,∴x22=4(y2+ 1)。
Again.
And ∵ y2 ≥-l, ∴ on = 2+y2.
Let the midpoint e of ON be perpendicular to the straight line passing through points n and e respectively, and the vertical feet are p and f, then,
∴ON=2EF，
That is, the distance from the midpoint of ON to a straight line is equal to half the length of ON.
∴ Inner diameter circle and tangent.
(3) If the intersection m is MH⊥NP and NP intersect at H, then,
∫y 1 = kx 1，y2=kx2，∴(y2-y 1)2 = k2(x2-x 1)2。 Mn2 = ( 1+K2) (x2-xl)2。
And ∵ points m and n are both on the image of y=kx and on the parabola.
∴, that is, x2-4kx-4 = 0, ∴ x2+x 1 = 4k, x2 x 1 =-4.
Mn2 =( 1+K2)(x2-XL)2 =( 1+K2)[(x2+XL)2-4x 2 XL]= 16( 1+K2)2。 ∴MN=4( 1+k2)。
Extend NP to point q, transfer point m to point s as MS⊥,
Then ms+NQ = y 1+2+y2+2 =
∴MS+NQ=MN, that is, the sum of the distances from two points to m and n is equal to the length MN of the line segment.
Example 6: (20 12 Xianning, Hubei 10) As shown in figure 1, in the rectangular MNPQ, points E, F, G and H are on NP, PQ, QM and MN respectively. If so, the quadrilateral EFGH is called the reflective quadrilateral of the rectangle MNPQ. Figures 2, 3 and 4.
Understanding and drawing:
(1) In Figure 2 and Figure 3, point E and point F are on the sides of BC and CD respectively. Try to make the reflection quadrilateral EFGH. Draw a rectangle ABCD on the diagram with a square grid. Calculation and guess:
(2) Find the perimeter of the reflective quadrilateral EFGH in Figure 2 and Figure 3, and guess whether the perimeter of the reflective quadrilateral of rectangular ABCD is constant. Enlightenment and proof:
(3) As shown in Figure 4, in order to prove the above conjecture, Xiaohua tried to extend the extension line from GF to BC to m, and tried to give it to me by Xiaohua.
Our inspiration proves the conjecture in (2).
The answer solution: (1) is drawn as follows:
(2) In Figure 2,
The perimeter of the quadrilateral EFGH is.
In fig. 3,
The perimeter of the quadrilateral EFGH is.
Guess: The perimeter of the reflecting quadrilateral of rectangular ABCD is a constant value.
(3) extending the extension line of GH intersection CB to point n,
∵,,∴。 ∶fc = fc，
∴Rt△FCE≌Rt△FCM(ASA)。 ∴EF=MF,EC=MC。
Similarly: NH=EH, NB=EB. ∴MN=2BC= 16。
∵,,,∴。 ∴GM=GN。
If point G is GK⊥BC in K, then.
∴。
The perimeter of the quadrilateral EFGH is. The perimeter of the reflecting quadrilateral of the rectangular ABCD is a constant.
Example 7: (20 12 Guangxi Chongzuo 10) As shown in the figure, in a square ABCD, points E and F move on BC and CD respectively, but the distance from point A to EF is always equal to the length of AB, which is during the movement of points E and F;
(1) ∠ Has the size of EAF changed? Please explain the reason.
(2) Has the circumference of △ ECF changed? Please explain the reason.
Second, the area (sum difference) is a fixed value problem:
Typical example: The copyright belongs to Jinyuan Mathematics Studio and cannot be reproduced.
Example 1: (20 12 Shiyan, Hubei, 3 points) As shown in the figure, O is a point within positive △ABC, OA=3, OB=4, OC=5. Take point B as the rotation center, rotate the line segment BO counterclockwise by 60 degrees to get the line segment BO'. The following conclusions are drawn: ①△BO'A can rotate counterclockwise around point B from△ △BOC; ② The distance between points O and O 'is 4; ③∠AOB = 150； ④; (5) The correct conclusion is that
A.①②③⑤ B.①②③④ C.①②③④⑤ D.①②③
Answer a.
The nature of central rotation, the judgment and nature of congruent triangles, the judgment and nature of equilateral triangle, the inverse theorem of Pythagorean theorem.
Analysis of ∵ positive △ABC, ∴AB=CB, ∠ABC=600.
∵ The line segment BO rotates 60 counterclockwise around point B to obtain line segments bo', ∴bo = bo', ∠ O' ao = 600.
∴∠o′ba=600-∠abo=∠oba。 ∴△bo′a≌△boc。
△BOC rotates 60 counterclockwise around point B to get ∴△BO'A, so conclusion ① is correct.
Connect OO',
∵BO=BO', ∠O'AO=600, ∴△OBO' are equilateral triangles. ∴oo′=ob=4。 Therefore, conclusion ② is correct.
∫ In △ AOO ′, the three sides are O ′ A = OC = 5, OO ′ = OB = 4, OA=3, which is a set of Pythagorean numbers.
∴△AOO' is a right triangle.
∴∠ AOB =∠ AOO'+∠ O' OB = 900+600 =150. Therefore, conclusion ③ is correct.
. So conclusion ④ is wrong.
As shown in the figure, rotate Δ△ AOB 60 counterclockwise around point A to make AB and AC coincide.
Point o rotates to point o ".
It is easy to know that △ AOO "is an equilateral triangle with a side length of 3 △ Chief Operating Officer" is a right triangle with a side length of 3, 4 and 5.
Then.
Therefore, conclusion ⑤ is correct.
To sum up, the correct conclusion is: 1235. So choose a.
Example 2: (20 12, Yulin, Fangchenggang, Guangxi, 12) As shown in the figure, in the plane rectangular coordinate system O, the coordinate of vertex A of rectangular AOCD is (0,4), and there are two moving points P and Q. Point P starts from point O and follows the OC line at a speed of 2 unit lengths per second (excluding endpoints O and C). D) Move to point d at a constant speed of/kloc-0 per unit length per second. Point p and point q start and stop at the same time. Let the exercise time be t seconds. When t=2 seconds, PQ=. (1) Find the coordinates of point D and directly write the range of t;
(2) Connect AQ, extend the intersecting axis to point E, extend AE along AD to point F, and connect EF. Does the area s of △AEF change with the change of t? If it changes, find the functional relationship between s and t; If it is the same, find the value of S. (3) Under the condition of (2), what is the value of T, and the quadrilateral APQF is trapezoidal?
Answer: (1) According to the meaning of the question, when t=2 (seconds), OP=4 and CQ=2.
In Rt△PCQ, from Pythagorean theorem, PC==4.
∴OC=OP+PC=4+4=8。 [Source: Zxxk.Com]
There are also: rectangular AOCD, a (0 0,4), ∴ d (8 8,4).
The range of t is: 0 < t < 4.
(2) Conclusion: The area s of △AEF remains unchanged.
* aocd is a rectangle, ∴AD∥OE, ∴△AQD∽△EQC.
∴, that is, the solution is CE=.
According to the nature of folding transformation, df = dq = 4-t, then cf = CD+df = 8-t.
S=S trapezoidal aocf+s △ FCE-s △ AOE = (OA+cf)? OC+CF？ CE-OA？ Old English
= [4+(8-t)]×8+(8-t)？ -×4×(8+)。
To simplify: S=32 is a constant value.
Therefore, the area s of △AEF remains unchanged, and S=32.
(3) If the quadrilateral APQF is trapezoidal, there is only PQ∑AF because AP and CF are not parallel.
It can be obtained from PQ∨AF:△CPQ∽△DAF.
∴ CP: AD = CQ: DF, that is, 8-2t: 8 = t: 4-t, which is simplified as T2- 12t+ 16 = 0.
Solution: t 1=6+2, t2=.
It can be seen from (1) that 0 < t < 4, ∴t 1=6+2 does not meet the meaning of the question, so it is discarded.
∴ When t= seconds, the quadrilateral APQF is a trapezoid. :Z*xx*k.Com]
Example 3: (20 12, Suzhou, Jiangsu, 9: 00) As shown in the figure, the side AD of the square ABCD coincides with the side FG of the rectangular EFGH, and the square ABCD moves in the FG direction at the speed of 1cm/s, and the point A coincides with the point F before the movement. During the movement, the side AD always coincides with the side FG and connects CG. When CG is at point P, it connects PD through the intersection line GH of parallel lines at point A.. It is known that the side length of square ABCD is 1cm, and the side lengths FG and GH of rectangular EFGH are 4cm and 3cm respectively. Let the moving time of the square be x(s) and the length of the line segment GP be y(cm), where 0≤x≤2.5.
(1) Try to find the functional relationship between y and x, and find the corresponding value of x when y =3;
(2) Note that the area of △DGP is S 1 and the area of △CDG is S2. Explain that S 1-S2 is a constant;
(3) When the straight line of the line segment PD is perpendicular to the diagonal AC of the square ABCD, find the length of the line segment PD.
Answer: (1)∵CG∨AP, ∴∠CGD=∠PAG, and then. ∴。
∵gf=4,cd=da= 1,af=x,∴gd=3-x,ag=4-x。
That's it. ∴y's functional relationship with x is.
When y =3, the solution is x=2.5.
(2)
∵,
∴ is a constant.
(3) Extend the intersection of PD and AC at Q point.
∫ In square ABCD, AC is diagonal, ∴∠ CAD = 45.
∵PQ⊥AC,∴∠ADQ=45 .∴∠GDP=∠ADQ=45 .
△ DGP is an isosceles right triangle, then GD=GP.
∴, simplified as:, solved as:.
∵0≤x≤2.5,∴。
At Rt△DGP,
Example 4: (20 12, Zigong, Sichuan, 12) As shown in the figure, in the rhombic ABCD, AB=4, ∠ Bad = 120, △AEF is a regular triangle, and points E and F slide on the side of BC. CD, e and f are different from B.
(1) proves that no matter how E and F slide on BC. CD, there is always Be = CF
(2) When point E and point F slide on BC. Whether the area of quadrilateral AECF and △CEF has changed is discussed respectively. If it is unchanged, find this constant value; If it changes, find the maximum (or minimum) value.
Answer: (1) proof: as shown in the figure, then communicate.
∫ quadrilateral ABCD is a diamond, ∠ bad = 120,
∠BAE+∠EAC=60，∠FAC+∠EAC=60 ,∴∠BAE=∠FAC。
∫∠ bad =120, ∴∠ ABF = 60. ∴△ ABC and △△△ ACD are equilateral triangles.
∴∠ACF=60，AC=AB .∴∠ABE=∠AFC。
∴, ∠∠BAE =∠fac, AB=AC, ∠ABE=∠AFC in △ABE and △ACF,
∴△ABE≌△ACF(ASA)。 ∴BE=CF。
(2) The area of quadrilateral AECF remains unchanged, while the area of △CEF changes. The reason for this is the following：
△ Abe △ ACF is obtained by (1), then S△ABE=S△ACF.
∴S quadrilateral aecf = s △ AEC+s △ ACF = s △ AEC+s △ Abe = s △ ABC, which is a constant value.
Let AH⊥BC be at point H, then BH=2,
According to the shortest vertical line segment, when the side AE of the regular triangle AEF is perpendicular to BC, the side AE is the shortest.
Therefore, the area of triangle AEF will change with the change of AE. When AE is the shortest, the area of regular triangle AEF is the smallest.
And S△CEF=S quadrilateral aecf-s △ AEF, then the area of △CEF will be the largest at this time.
∴S△CEF=S quadrilateral aecf.
∴△△△ What is the maximum area of CEF?
Example 5: (20 12, Yiyang, Hunan, 12) is known as figure 1. In a square ABCD with an area of 3, e and f are two points on the side of BC and CD, AE⊥BF is at point G, and BE = 1.
(1) verification: △ Abe △ BCF;
(2) Find the area of the overlapping part of △ABE and △BCF (i.e. △ beg);
(3) Now rotate △ABE to △ AB ′ E ′ counterclockwise around point A (as shown in Figure 2), so that point E falls on point E ′ on the edge of CD. Ask whether the area of the overlapping part of △ABE and △BCF has changed before and after rotation. Please explain the reason.
The answer (1) proves that ∵ quadrilateral ABCD is a square, ∴∠ Abe = ∠ BCF = 90, AB=BC. ∴∠ABF+∠CBF=90 .
∵AE⊥BF,∴∠ABF+∠BAE=90 .∴∠BAE=∠CBF。
In △ABE and △BCF, ∠∠ABE =∠BCF, AB=BC, ∠BAE=∠CBF,
∴△ABE≌△BCF(ASA)。
(2) Solution: ∫ Square area is 3, ∴AB=.
In △BGE and △ABE, ∠∠gbe =∠BAE, ∠ EGB = ∠ EBA = 90, ∴△BGE∽△ABE.
∴。
∫be = 1，∴AE2=AB2+BE2=3+ 1=4.∴。
Third, other valuation issues: the copyright belongs to Jinyuan Mathematics Studio and cannot be reproduced.
Typical example: The copyright belongs to Jinyuan Mathematics Studio and cannot be reproduced.
Example 1: (20 12 points, Yiwu, Zhejiang 12 points) As shown in figure 1, it is known that the straight line y=kx intersects the parabola at point A (3,6). (1) Find the analytical formula of straight line y=kx and line segment length OA;
(2) Point P is the moving point in the first quadrant of the parabola, the intersection point P is a straight line PM, the X axis is at point M (points M and O are not coincident), the straight line OA is at point Q, then point Q is the vertical line of the straight line PM, and the Y axis is at point N. Try to explore: Is the ratio of the length of the straight line QM to the straight line QN constant? If yes, find out the fixed value; If not, explain the reasons;
(3) As shown in Figure 2, if point B is the point on the right side of the symmetry axis on the parabola, point E is on the line segment OA (not coincident with point O and point A), and point D (m, 0) is the moving point on the positive semi-axis of the X axis, which satisfies ∠ BAE = ∠ Bed = ∠ AOD. Continue to explore: when m is in what interval, meet the conditions.
Solution: (1) substitute point a (3,6) into y = kx6=3k, that is, k=2.
∴y=2x。 ∴。
(2) The ratio of the length QM of the line to the length QN of the line is constant for the following reasons:
As shown in figure 1, the QG⊥y axis is at G point, and the QH⊥x axis is at H point.
① When QH and QM overlap, it is obvious that QG and QN overlap.
At this moment.
② When QH and QM do not coincide,
Qg⊥qh ∵qn⊥qm, it is suggested to set H point and G point on the positive semi-axis of X axis and Y axis respectively.
∴∠MQH=∠GQN。
∠∠qhm =∠qgn = 90，∴△QHM∽△QGN.∴。
The same principle can be obtained when the positions of point P and point Q are different on parabola and straight line.
The ratio of the length of line QM and QN is a constant.
(3) As shown in Figure 2, extend the intersection X axis of AB at point F, FC⊥OA at point C and point F, and AR⊥x axis at point R and point A ... The copyright belongs to Jinyuan Mathematics Studio and cannot be reproduced.
∵∠AOD=∠BAE,∴AF=OF。 ∴OC=AC=。
∠∠ARO =∠FCO = 90，∠AOR=∠FOC，
∴△AOR∽△FOC。 ∴。 ∴OF=。 ∴ point f (0).
Let point b (x,) pass through point b as BK⊥AR of point k, and then △AKB∽△ARF.
That's it.
The solution is x 1=6, x2=3 (truncation). ∴ point b (6 6,2).
∴BK=6﹣3=3,AK=6﹣2=4。 ∴AB=5。
In △ABE and △OED, ∠∠BAE =∠bed, ∴∠ Abe+∠ AEB = ∠ deo+∠ AEB. ∴∠ Abe = ∠.
∵∠BAE=∠EOD,∴△ABE∽△OED。
Let OE=x, AE =-x (),
From △ABE∽△OED, that is.
∴。
The vertex is.
As shown in Figure 3, OE=x= at that time, and there are 1 points at this time;
At this time, any value of m corresponds to two values of x, and at this time, e has two points.
∴ At that time, point E was only 1 point. At that time, point E was 2 points. ..
Example 2: (Shandong Zibo 20 12, 4 points) As shown in the figure, the square is folded in half and then unfolded (Figure ④ is folded twice in succession and then unfolded), and then a right triangle can be obtained by the method shown in the figure, one right angle of which is equal to half of the hypotenuse. These figures are as follows.
(A)4 (B)3 (C)2 (D) 1。
Answer C.
Test the properties of central square, folding, right triangle with 30-degree angle, parallelism, judgment of isosceles triangle, midline on hypotenuse of right triangle, and triangle interior angle theorem.
The analysis is shown in figure 1, ∠ ABC = ∠ Abd.
Namely ∠ ABC < 22.50.
According to the property that the right side of a right triangle with an angle of 30 degrees is half of the hypotenuse, CD≠BC.
In Figure 2, according to the nature of folding, ∠ABC=∠ABF, EC∨FB,
∴∠ABC=∠ABF=∠ADE=∠BDC。 ∴BC=DC。
There is also the nature of folding a square in half and the nature of parallel lines, knowing that AD=BD,
According to the nature of the midline on the hypotenuse of a right triangle, DC=AB, that is, BC=AB.
Satisfy that the right angle side is equal to half of the hypotenuse.
The nature of the square in fig. 3 is that one right-angled side is equal to half of the other right-angled side, and it is impossible for another right-angled side to be equal to half of the hypotenuse.
In fig. ④, it can be seen from the properties of square folding and parallel lines that AB=CB, AB=2BD,
∠ABE=∠CBE，
∴BC=2BD。 ∴∠BCD=300。 ∴∠CBD=600。
∠∠ABE+∠CBE+∠CBD = 1800。 ∴∠ABE =600 .∴∠AEB =300 .
∴AB=BE。 Satisfy that the right angle side is equal to half of the hypotenuse.
To sum up, there are two such diagrams. So choose C.
Example 3: (20 12 Mianyang, Sichuan 14) As shown in figure 1, in the rectangular coordinate system, O is the coordinate origin, point A is on the positive semi-axis of Y axis, and the image F of quadratic function y=ax2+x +c intersects with X axis at points B AM=BC, and intersects with Y axis at point M. (1) Find the analytic formula of quadratic function; The copyright belongs to Jinyuan Mathematics Studio and cannot be reproduced.
(2) Prove that there is a point D on the parabola F, so that the quadrilateral formed by connecting the four points A, B, C and D happens to be a parallelogram, and find the analytical formula of the straight line BD;
(3) Under the condition of (2), let the straight line L cross D, intersect with the straight lines BA and BC at different points P and Q, and AC and BD intersect at N. ..
(1) if the straight line l⊥BD, as shown in figure 1, try to find the value;
② If l is an arbitrary straight line that meets the conditions. As shown in Figure 2, is the conclusion in ① still valid? If so, prove your guess; If not, please give a counterexample.
Solution to the answer: (1)∫ The image of quadratic function y=ax2+x +c passes through b (-3,0), M (0, 1),
∴, solution.
∴ The analytic formula of quadratic function is:
(2) Proof: Let y=0, and X 1 =-3 and x2=2 are obtained.
∴C(2,0),∴BC=5。
Let x=0, y=- 1, ∴M(0,- 1), OM= 1.
And AM=BC, ∴ OA = am-om = 4. ∴A(0,4)。
Let the AD∑x axis intersect the parabola at point D, as shown in figure 1.
Then, the solution is x 1=5, X2 =-6 (in the second quadrant, omitted).
The coordinate of point d is (5,4). ∴AD=BC=5。
And ∵ad∨BC, ∴ quadrilateral ABCD is a parallelogram, that is, there is a point D on parabola F, so the quadrilateral formed by connecting four points A, B, C and D happens to be a parallelogram. The copyright belongs to Jinyuan Mathematics Studio and cannot be reproduced.
Let the analytical formula of straight line BD be: y=kx+b, ∫ b (-3,0), d (5 5,4),
∴, solution:. ∴ The analytical formula of straight BD is:
(3) In Rt△AOB,
AD=BC=5，∴？ ABCD is a diamond.
① If the straight line l⊥BD is shown in figure 1,
∵ quadrilateral ABCD is a diamond, ∴AC⊥BD.∴AC∥ straight line ∴.
∵BA=BC=5,∴BP=BQ= 10。
∴。
② If L is an arbitrary straight line that meets the conditions, as shown in Figure 2, the conclusion in ① still holds for the following reasons:
∵AD∥BC,CD∥AB,∴△PAD∽△DCQ。 ∴。
∴AP？ CQ = AD? CD=5×5=25 .
∴
Example 4: (20 12, Chengdu, Sichuan, 12) As shown in the figure, in the plane rectangular coordinate system xOy, the image of a linear function (constant) intersects the X axis at point A (0) and the Y axis at point C, and a parabola (a) with the straight line x= 1
(1) and parabolic function expressions;
(2) Let E be a point on the parabola on the right side of the Y axis, and the parallel line passing through E is a straight line AC intersecting with the X axis at point F. Is there such a point E, and the quadrilateral with vertices A, C, E and F is a parallelogram? If it exists, find the coordinates of point E and the area of the corresponding parallelogram; If it does not exist, please explain the reason;
(3) If p is the point on the parabola symmetry axis that minimizes the circumference of △ACP, make a straight line that is not parallel to the Y axis and intersect with two points of parabola, try to find out whether it is a fixed value, and write the inquiry process.
Answer solution: (1)∵ crossing point (3,0), ∴, solution.
The analytical formula of a straight line is.
Let x=0, so. ∴C(0,)。
∵ parabola y=ax2+bx+c, the symmetry axis is x= 1, it intersects the x axis at a (∵ 3,0), and the other intersection point is B (5 5,0).
Let the analytical formula of parabola be y = a (x+3) (x < 5),
∵ The parabola passes through C(0,), ∴=a? 3 (-5), get the solution.
The analytical formula of parabola is y = (x+3) (x-5), that is.
(2) Suppose there is a point E that the quadrilateral with vertices A, C, E and F is a parallelogram.
Then AC∑EF and AC=EF, as shown in figure 1.
(i) When the point E is at the position of the point E, the EG⊥x axis passing through the point E is at the point G,
∵AC∥EF,∴∠CAO=∠EFG。
∵∠COA=∠EOF=900，AC=EF，
∴△CAO≌△EFG(AAS)。
, that is, leaf =
∴,
Xe = 2 is obtained (Xe = 0 coincides with point C and is discarded).
∴E(2,),S？ ACEF= .
(ii) When the point E is at the position of the point e', the point e' intersects with the point g' because the E'G'⊥x axis is at the point G',
Similarly, we can get e' (), s? ACE ' F ' = .
(3) To minimize the perimeter of △ACP, it is only necessary to minimize AP+CP.
As shown in Figure 2, BC is connected with x= 1 at point P, because point A and point B are symmetrical about x= 1. According to the axisymmetric property and the shortest line segment between two points, we can know that AP+CP is the minimum at this time (the minimum value of AP+CP is the length of line segment BC).
∫B(5，0)，C(0，)，
The analytical formula of line BC is.
∫XP = 1, ∴yP=3, which is P (1, 3).
Let the straight line passing through point p (1, 3) be y = kx+3-k,
Lian lide
x2+(4k﹣2)x﹣4k﹣3=0，
∴x 1+x2=2﹣4k,x 1x2=﹣4k﹣3。
∵y 1=kx 1+3﹣k,y2=kx2+3﹣k,∴y 1﹣y2=k(x 1﹣x2)。
According to Pythagorean theorem:
M 1M2=
M 1P=，
M2P= .
∴M 1P？ M2P=
∴M 1P？ M2P=M 1M2 .∴= 1 is a fixed value.