Junior high school mathematics is a difficult ending.

1, as shown in figure 1, it is known that the side length of a square OABC is 2, the vertices A and C are on the positive semi-axes of the X and Y axes respectively, m is the midpoint of BC, P(0, m) is a moving point on the line segment OC (except point C), and the extension line of the straight line PM intersecting AB is at point D. 。

(1) Find the coordinates of point D (expressed by an algebraic expression containing m);

(2) When △APD is an isosceles triangle, find the value of m;

(3) Let the parabola passing through P, M and B intersect with the positive semi-axis of the X axis at point E, the passing point O is the perpendicular of the straight line ME, and the vertical foot is H (as shown in Figure 2). When point P moves from point O to point C, point H moves with it. Please write down the path length of point H directly (without writing the solution process).

2. As shown in the figure, in the diamond-shaped ABCD, AB = 10, Sina = 45, point E is on AB, AE = 4, the intersection point E is EF∨AD, the CD intersects at point F, point P starts from point A, and moves along the AB line at a constant speed of 1 unit per second to the end point B, and point Q starts from E.

(1) When t = 5 seconds, find the length of PQ;

(2) When BQ bisects ∠ABC, the straight line PQ divides the circumference of the diamond-shaped ABCD into two parts, and finds the ratio of these two parts;

(3) Can ⊙P with P as the center and PQ length as the radius be tangent to the straight line AD? If yes, find the value of t at this time; If not, explain why.

Solution: (1) Draw a picture according to the meaning of the question, as shown in the figure: point P is PM⊥EF, and vertical foot is M.

If AE=4 and AP=EQ=5, then EP= 1.

∫EF∨AD，

∴∠BEF=∠A, that is, sin∠BEF=sinA=4/ 5.

That is, PM EP =4/ 5, then PM=4/ 5,

According to Pythagorean theorem, EM=3 /5,

Then MQ=5-3/ 5 =22/ 5,

In the right triangle of PQM, according to Pythagorean theorem:

PQ =(4 ^ 5)2+(22 ^ 5)2 = 2 ^ 5；

(2) Draw a picture according to the meaning of the question, as shown:

∫BQ average score ∠ABC,

∴∠EBQ=∠CBQ，

And ∵BC∨EF,

∴∠CBQ=∠EQB，

∴∠EBQ=∠EQB，

∴EB=EQ= 10-4=6，

T=6，AP=6，

∴BP=4,QF=4，

Let PQ cross CD at point m,

∫AB∨CD，

∴∠EPQ=∠FMQ,∠PEQ=∠MFQ，

∴△EPQ∽△FMQ，

∴EP/ FM =EQ/ QF, that is, 2 /FM =6 /4,

∴FM=4 /3，

Then MD=4-4/ 3 =8 /3, MC=22 /3,

Then the perimeters of the two parts of the straight line PM divided into diamonds are AP+AD+MD and PB+BC+CM, respectively.

That is, the circumference of a diamond is divided into 56 /3 and 64 /3.

So the ratio of these two parts is 7: 8;

(3) passing through P at H to reach PH⊥AD, passing through EF at G,

Then PH=4 /5 t, PE=t-4, PG=4/ 5 (t-4), EG=3/ 5 (t-4),

∴GQ=t-EG=2/ 5 t+ 12 /5，

PQ2 = PG2+GQ2 =(4/5t- 16/5)2+(2/5t+ 12/5)2，

Equation (4/5t) 2 = (4/5t-16/5) 2+(2/5t+12/5) 2,

Solution: t = 10.

3. It is known that the quadrilateral ABCD is a square with a side length of 4, with AB as the diameter, and a semicircle is made in the square. P is the moving point on the semicircle (not coincident with point A and point B), which connects PA, PB, PC and PD.

(1) As shown in Figure ①, when the length of PA is equal to _ _ _ _ _ _ _, ∠ PAB = 60;

When the length of PA is equal to _ _ _ _ _ _ _, △PAD is an isosceles triangle;

(2) As shown in Figure ②, take the straight line on AB side as the X axis and the straight line on AD side as the Y axis, and establish a rectangular coordinate system as shown in the figure (point A is the origin O), and the areas of △PAD, △PAB and △PBC are S 1, S2 and S3 respectively. Let the coordinates of point P be (a, b), and try to find 2S 1S3.

4. As shown in the figure, in the rectangular ABCD, AB = 4, AD = 2, point M is the midpoint of AD, and point E is the moving point on the side of AB. Connect EM, extend the intersecting ray CD at point F, the perpendicular line intersecting M, such as EF and BC, intersect at point G, connect EG, and intersect with DC at point H. Let the length of AE be x, and the area of △MEG be y. 。

(1) Find the value of sin∠MEG;

(2) Find the resolution function of Y with respect to X, and determine the range of the independent variable X;

(3) Let the midpoint of line segment MG be n and connect CN. Is there a value of x that makes a triangle with vertices n, c and g similar to △EFH? If it exists, find the values of x and y; If it does not exist, please explain why.

Solution: (1) If the intersection point G is GN⊥AD and the extension line of AD is at point N, it can be proved that △AEM∽△NMG.

∴MG /EM =GN/ MA，

∴GN=AB=4，

∫M is the midpoint of AD,

∴AM= 1，

∴MG/ EM =GN/ MA =4，

∵GM⊥EF，

Ⅶ in Rt△EMG

∴tan∠meg=mg/em = 4；

(2) From (1), it is known that MG /EM =4, that is, MG=4EM,

∫ In Rt△AEM, EM= x2+ 1,

∴MG=4 x2+ 1，