Exploring the changing law of geometric figures from the perspective of motion is called dynamic geometric questions, and the resulting dynamic geometric questions are research. In the movement of geometric figures, the relationship between figure position and quantity has certain "change" and "unchanged" questions. As far as moving objects are concerned, there are some motions, linear motions and plane motions, and as far as their motion forms are concerned, there are translation, rotation, folding and rolling.
Dynamic geometry test questions are flexible, dynamic and static, which can develop students' spatial imagination and comprehensive analysis ability in the process of movement changes and become a hot spot in the senior high school entrance examination in recent years. Here, taking the 2006 and 2007 senior high school entrance examination questions as examples, the dynamic geometry questions are classified and analyzed.
Question 1: inching type
The inching formula is to design one or several moving points on some geometric figures such as triangle, rectangle and trapezoid, and to study the equivalence relationship, variable relationship, special state of the figure and special relationship between the figures produced by these points in the process of motion change.
1. Single acting point type
Example1(Twelve Cities in Liaoning Province in 2007) As shown in the figure, it is known that in the equilateral triangle ABC, points D, E and F are the midpoints of sides AB, AC and BC respectively, m is the moving point on the straight line BC, and △DMN is an equilateral triangle (when the position of point M changes, △DMN also moves as a whole).
(1) As shown in Figure ①, when point M is to the left of point B, please judge the quantitative relationship between en and MF. Is point f on a straight liNE ne? Please write the conclusion directly, without proof or explanation;
(2) As shown in Figure ②, when point M is on BC, other conditions remain unchanged. (1) Is the quantitative relationship between en and MF still valid? If yes, please use Figure ② to prove it; If not, please explain the reasons;
(3) If point M is to the right of point C, please draw the corresponding figure in Figure ③ to judge whether the quantitative relationship between en and MF in the conclusion (1) still holds. What if it is established? Please write the conclusion directly, without proof or explanation.
Exercise 1 (Fuzhou, 2007) As shown in the figure, the straight line AC‖BD, the connecting line AB, the straight line AC, BD, and the line segment AB divide the plane into four parts, stipulating that all points on the line do not belong to any part. When the moving point p falls on a certain part, connect PA and PB to form ∠PAC.
(1) When the fixed point p falls in the first part, it is proved that: ∠ APB = ∠ PAC+∠ PBD;
(2) Does ∠APB =∠PAC +∠PBD hold when the moving point P falls in the second part (directly answer yes or no)?
(3) When the moving point P falls in the third part, fully explore the relationship between ∠PAC, ∠APB and ∠PBD, and write the specific position of the moving point and the corresponding conclusions. Choose one of the conclusions to prove.
Exercise 2 (Mianyang City, 2006) In the square ABCD, point P is a moving point on the CD, connecting PA, and passing through point B and point D respectively are BE⊥PA and DF⊥PA, with steps of E and F respectively, as shown in Figure ①.
(1) Please explore the quantitative relationship BEtween the lengths of be, DF and EF. If point P is on the extension line of DC (Figure ②), what is the quantitative relationship between the lengths of these three line segments? What if point P is on the extension line of the CD (as shown in Figure ③)? Please write the conclusion directly;
(2) Please choose one of the three conclusions in (1) to prove it.
The analogy discovery method is often used to solve this kind of moving point geometry problems, that is, by observing and comparing the similarities and differences between two or several similar mathematical research objects, starting with the nature of one easy-to-explore research object, guessing the similar nature of another or several similar figures, and then drawing relevant conclusions. Analogy discovery can roughly follow the following steps: (1) According to known conditions, analyze and observe possible situations from a dynamic perspective. (2) Combined with the corresponding figures, with static braking, relevant conclusions are drawn by using the learned knowledge (such as triangle congruence and triangle similarity). ). (3) Guess the nature of graphics in other cases by analogy.
2. Double moving point type
Example 2 (Listening to Harbin in 2007) As shown in figure 1, in the square ABCD, the diagonal AC and BD intersect at point E, and the bisector AF ∠BAC and BD intersect at point F. 。
(1) Verification:;
(2) point C 1 starts from point c and moves to point b along the line segment CB (not coincident with point b). At the same time, point A 1 starts from point A and moves along the extension line of BA at the same speed. When the moving point stops moving, another moving point A 1 also stops moving. As shown in figure 2, A 1F655. Crossing BD at point F 1, crossing point f1e 1 ⊥ a1,and the vertical foot is e1.Please guess e1f/kloc-0.
(3) under the condition of (2), when A65438+E 1 = 3 and C65438+E 1 = 2, find the length of BD.
3. Multi-moving point type
Example 3 (Meishan City, 2006) is as follows: ∠ mON = 90°, there is a square AOCD inside ∠MON, points A and C are ON ray OM and ON respectively, point B 1 is any point on ON, and a square AB/kloc-0 is made inside ∠MON.
(1) connect D 1D, and verify: ∞∠ add1= 90; ;
(2) Connect CC 1 and guess what the degree of ∠C 1CN is? And prove your conclusion;
(3) Take any point B2 ON the ON, take AB2 as the edge, make a square AB2C2D2 in ∠MON, observe the figure, and combine the conclusions of (1) and (2), please make a reasonable judgment again.
The drill (Yichang City, 2007) is shown in figure 1. In △ABC, AB = BC = 5 and AC=6. △ECD is obtained by translating △ABC in BC direction and connecting AE, AC and BE to intersect at point O. 。
(1) Determine what a quadrilateral ABCE is and explain the reasons;
(2) As shown in Figure 2, P is a moving point on the BC line (Figure 2), which is not coincident with points B and C), connects PO and extends the intersection AE, QR⊥BD at point Q, and the vertical foot is point R. 。
① Does the area of quadrilateral PQED change with the movement of point P? If yes, please explain the reasons; If not, find the area of quadrilateral PQED;
② When the length of line segment BP is what value, is △PQR similar to △BOC?
Through the above example, we can find that the double moving point problem can be transformed into a single moving point problem, and the key is to grasp the key moving point that determines the whole problem, thus transforming the problem.
Question 2: Linetype
1. Line conversion type
Example 4 (Leshan City, 2007) shows that in the rectangular ABCD, AB=4, AD = 10. When the right-angle vertex P of the square ruler slides on AD (point P does not coincide with points A and D), the right-angle side passes through point C, and the other right-angle side AB intersects with point E, we know that the conclusion is "RT △ AEP ∽".
(1) When ∠ CPD = 30, find the length of AE;
(2) Is there such a point p that the perimeter of △DPC is equal to twice the perimeter of △AEP? If yes, find out the length of DP; If it does not exist, please explain why.
2. Line rotation type
Example 5 (Hengyang City, 2006) is known: As shown in the figure, in the parallelogram ABCD, AB⊥AC, AB= 1, diagonal AC and BD intersect at point O, and the straight line AC, BC and AD rotate clockwise around point O intersect at points E and F respectively.
(1) proves that the quadrilateral ABEF is a parallelogram when the rotation angle is 90;
(2) Try to explain that in the process of rotation, the line segment AF and EC always remain equal;
(3) Could the quadrangular BEDF be rhombic in the process of rotation? If not, please explain the reasons; If yes, explain the reason and find out the degree of clockwise rotation of AC around point O at this time.
The essence of linear motion is inching, that is, inching drives linear motion, and then surface motion will occur. Therefore, the geometric problem of linear motion can be transformed into a point motion problem to solve. The key to solve this kind of problem is to grasp the whole process of graphic movement and change, and grasp the equal relationship and variable relationship between them. From the movement of the figure to the special position, we can explore the general conclusion or get inspiration from it. This idea from special to general is extremely important for us to solve the problem of motion change.
Question 3: There are three basic transformations in the movement transformation of graphics: translation, rotation and folding. It is mainly to change the position of a given figure (or part of it), and then analyze the relationship between the figures in the new figure. This kind of questions often combine inquiry and existence to examine students' practical ability, observation ability, exploration and practical ability.
1. Graphic translation type
Example 6 (Hebei, 2007) ABC, AB=AC, intersects with the extension line of BA at point G. As shown in figure 1, put an isosceles right-angled triangular ruler, whose right-angled vertex is F, one right-angled side is in a straight line with AC side, and the other right-angled side just passes through point B. 。
(1) In figure 1, please guess and write the quantitative relationship between BF and CG by observing and measuring the length of BF and CG, and then prove your guess;
(2) When the triangular ruler is translated to the position shown in Figure 2 along the AC direction, one right-angle side is still on the same line with the AC side, and the other right-angle side intersects with the BC side at point D, and the intersection point D is DE⊥BA at point E. At this time, please guess and write the quantitative relationship between DE+DF and CG by observing and measuring the lengths of DE, DF and CG, and then prove your guess;
(3) When the trigonometric ruler continues to translate along the AC direction to the position shown in Figure 3 on the basis of (2) (point F is on the line segment AC, and point F does not coincide with point C), does the conjecture in (2) still hold? (No need to explain why)
The translation of graphics is essentially the translation of straight lines, which will produce similar graphics, so the key idea to solve this kind of problem is to get the relationship between the quantities to be solved through similarity. This topic is designed with a triangle as the background. When solving, we must understand the characteristics of triangles to reduce the difficulty of solving. Through solving, we can also see that triangle can change many wonderful math problems in the senior high school entrance examination through proper operation, which have appeared frequently in the senior high school entrance examination in the past two years.
2. Graphic rotation type
Example 7 (Linyi City, 2007)
As shown in figure 1, it is known that in △ABC, AB = BC = 1, ∠ ABC = 90. Put the right-angled vertex D of a triangle deF with an angle of 30 on the midpoint of AC (the short right-angled side of the triangle is DE and the long right-angled side is DF), and rotate the right-angled triangle DEF counterclockwise around the point D. 。
(1) In figure 1, DE crosses AB in M and DF crosses BC in N. 。
① Prove that DM = DN
② In this process, the overlapping part of right triangle DEF and △ABC is a quadrilateral DMBN. Please explain whether the area of the quadrangle DMBN has changed. If yes, please explain how? If not, calculate its area;
⑵ Continue to rotate to the position shown in Figure 2, and extend AB intersection point DE to M, BC intersection point DF to N, and DM = DN. Is it still valid? If yes, please give proof; If not, please explain the reasons;
(3) Continue to rotate to the position shown in Figure 3. Is FD across BC to N, ED across AB to M, and DM = DN still valid? If yes, please write an unproven conclusion.
Exercise 1 (Changde City, 2006) Overlaps two congruent right-angled triangles ABC and DEF, so that the acute vertex D of the triangle DEF coincides with the hypotenuse midpoint O of the triangle ABC, where ∠ ABC = ∠ DEF = 90, ∠ C = ∠ F = 45 and AB =
(1) As shown in figure 1, it is easy to prove that when ray DF passes through point B, △ APD ∽△ CDQ, that is, point Q coincides with point B, and at this time, APCQ =;
(2) rotate the triangle DEF counterclockwise around the o point from the position shown in fig. 9, and let the rotation angle be α, where 0.