1. Quadratic function and the shape of quadrilateral

Example 1. (Yiwu, Zhejiang Province) As shown in the figure, parabola and X axis intersect at point A and point B (point A is on the left of point B), and straight line and parabola intersect at point A and point C, where the abscissa of point C is 2.

(1) Find the coordinates of point A and point B and the functional expression of straight line AC;

(2)P is the moving point on the line segment AC, and the point passing through P is the plane of the Y axis.

The line intersects the parabola at point E, and the maximum PE length of the line segment is found.

(3) Point G is a moving point on a parabola. Is there a point F on the X axis, which makes the quadrilateral with four points A, C, F and G as its vertices a parallelogram? If it exists, find out all the coordinates of F points that meet the conditions; If it does not exist, please explain why.

Exercise 1. (Henan Experimental Zone) 23. As shown in the figure, a parabola with a straight axis of symmetry passes through this point.

A (6 6,0) and B (0 0,4).

(1) Find the parabolic analytical formula and vertex coordinates;

(2) Let point E (,) be the moving point on the parabola, located in the fourth quadrant, and the quadrilateral OEAF is a parallelogram with OA as the diagonal. Find the functional relationship of the area s sum of parallelogram OEAF, and write the range of independent variables.

① When the area of the parallelogram OEAF is 24, please judge whether the parallelogram OEAF is a diamond?

② Is there a point E that makes the parallelogram OEAF square? If it exists, find the coordinates of point e; If it does not exist, please explain why.

Exercise 2. Deyang City, Sichuan Province, 25. As shown in the figure, it is known that the vertex of the parabola that intersects the axis at points and is, the parabola is symmetrical with the axis, and the vertex is.

(1) Find the function relation of parabola;

(2) It is known that the origin, the fixed point, the point on the plane and the point on the plane are always symmetrical, so when the point moves, the quadrilateral with the point as the vertex is a parallelogram?

(3) Is there anything in the world that makes it a right triangle with a hypotenuse and an angle? If it exists, find the coordinates of the point; If it does not exist, explain why.

Exercise 3. (Shanxi Volume) As shown in the figure, it is known that the intersection of parabola and coordinate axis is,,.

(1) Find the analytical formula of parabola symmetrical about the origin;

(2) Let the vertex of the parabola be, the parabola and the axis intersect at two points respectively (the point is on the left side of the point), the vertex be, and the area of the quadrilateral be. If it is a point, the point will move left and right in the horizontal direction at the speed of 1 unit per second; At the same time, the point and the point move up and down in a fixed direction at a speed of 2 units per second until the point and the point coincide. Find the relationship between the area of quadrilateral and the movement time, and write the range of independent variables.

(3) When the value is, the area of the quadrilateral has a maximum value, and find this maximum value;

(4) Can a quadrilateral form a rectangle in the process of movement? If yes, find the value at this time; If not, please explain why.

2. Quadratic function and the area of quadrilateral

Example 1. (Ziyang) 25 As shown in figure 10, it is known that the parabola P: Y = AX2+BX+C (A ≠ 0) intersects with the X-axis at points A and B (point A is on the positive semi-axis of the X-axis), intersects with the Y-axis at point C, and one siDE de of the rectangle deFG is on the line segment AB.

x…-3-2 1 2…

y…-

-4 -

0 …

(1) Find the coordinates of points A, B and C;

(2) If the coordinate of point D is (m, 0) and the area of rectangular DEFG is S, find the functional relationship between S and M, and point out the range of M;

(3) When the area S of the right-angle DEFG is the largest, connect DF and extend to point M, so that FM=k? DF, if point m is not on parabola p, find the range of k.

Exercise 1. (Question 26, Twelve Cities in Liaoning Province, 2007). As shown in the figure, there is a right-angled trapezoid OMNH in the plane rectangular coordinate system, and its coordinates are H point (-8,0) and N point (-6,4).

(1) Draw a graph OABC in which the right-angled trapezoid OMNH rotates180 around point O, and write the coordinates of vertices A, B and C (point M corresponds to point A, point N corresponds to point B and point H corresponds to point C);

(2) Find the expression of parabola passing through points A, B and C;

(3) Intercept CE=OF=AG=m, E, F and G are on line segments CO, OA and AB, respectively. Find the functional relationship between quadrilateral BEFG area S and M, and write the range of independent variable M; Is there a minimum area s? If it exists, request this minimum value; If it does not exist, please explain the reason;

(4) In the case of (3), whether the adjacent sides of the quadrilateral BEFG are equal, if so, please directly write the value of m at this time and point out the equal adjacent sides; If it does not exist, explain why.

Exercise 3. As shown in the figure, the side length of a square is, and there is a nail moving point in the center of symmetry. At the same time, starting from a point, the point moves at a speed of every second in the direction and stops at the point, and the point moves at a speed of every second in the direction and stops at the point. The two points are connected by a flexible thin rubber band, and the area swept by the rubber band after seconds is.

(1) When, the functional relationship between summation;

(2) When the rubber band just touches the nail, evaluate it;

(3) When, the function relation of summation, and the change range of the rubber band from hitting the nail to stopping the movement;

(4) When, please draw the function image between and in the given rectangular coordinate system.

Exercise 4. (Sichuan Ziyang Volume) As shown in the figure, it is known that the image of parabola l 1: y = x2-4 intersects with X axis at two points A and C, B is the moving point of parabola l 1 (B does not coincide with A and C), parabolas l2 and l 1 are symmetrical about X, and the parallelogram ABCD is the fourth diagonal of AC.

(1) The analytical formula for finding l2;

(2) Verification: Point D must be on l2;

(Can ABCD be rectangular? If it can be a rectangle, find the area of the common part of these rectangles (if only one rectangle meets the conditions, find the area of this rectangle); If it cannot be rectangular, please explain why. Note: The calculation result is not approximate.

.

3. Dynamic exploration of quadratic function and quadrilateral.

Example 1. (Jingmen City) 28. As shown in figure 1, in the plane rectangular coordinate system, there is a rectangular piece of paper OABC, which is known as O (0 0,0), A (4 4,0), C (0 0,3), and point P is the moving point on the edge of OA (not coincident with points O and A). Now, choose a suitable point E on the edge of OC, and fold △POE along PE to get △PFE, so that the straight lines PD and PF coincide.

(1) Let P(x, 0) and E(0, y), find the functional relationship between y and x, and find the maximum value of y;

(2) As shown in Figure 2, if the folding point D falls on the edge of BC, find the parabolic function relationship of points P, B and E;

(3) In the case of (2), is there a point Q on the parabola that makes △PEQ a right triangle with PE on the right? If it does not exist, explain the reasons; If it exists, find the coordinates of Q point.

Example 2. (20 10 Shenyang No.26) It is known that the parabola Y = AX2+BX+C intersects with the X axis at points A and B, and intersects with the Y axis at point C, where point B is on the positive semi-axis of the X axis and point C is on the positive semi-axis of the Y axis, and the line segment lengths OB and OC (OB

(1) Find the coordinates of points A, B and C;

(2) Find the expression of this parabola;

(3) connect AC and BC. If point E is the moving point on the AB line (not coincident with points A and B), point E is the intersection of EF∑AC and BC at point F, then CE, let AE be m in length and S in area △CEF, find the functional relationship between S and M, and write the range of independent variable M;

(4) On the basis of (3), try to explain whether there is a maximum value of S, if so, request the maximum value of S, find the coordinates of point E at this time, and judge the shape of △BCE at this time; If it does not exist, please explain why.

Example 3 .. (Chenzhou, Hunan) 27. As shown in the figure, in the rectangle ABCD, AB = 3, BC = 4, the rectangle ABCD translates along diagonal A, and the translated rectangle is EFGH(A, E, C, G are always on the same straight line), and stops moving when points E and C are heavier. In translation, EF and BC intersect at n point, and the extension line of GH and BC intersects at m point.

Is (1) S equal to? Please explain the reason.

(2) Let AE = X, write the functional relationship between S and X, and find out what value X takes when S has the maximum value, and what is the maximum value?

(3) As shown in figure 1 1, when AE is of what value, the connecting BE is an isosceles triangle.

Exercise 1. (Hechi City, 2007) As shown in figure 12, the quadrilateral OABC is a right-angled trapezoid with A (4 4,0), B (3 3,4) and C (0 0,4). The pointer moves at a speed of 2 unit lengths per second; From the same time, these points move at a speed of/kloc-0 per unit length per second. When one moving point reaches the end point, another moving point stops moving. The intersection is perpendicular to the point, connecting AC with NP and MQ.

(1) point (fill in m or n) to reach the finish line;

(2) Find the functional relationship between the area s of △AQM and the movement time t, and write it by the following formula.

The range of the variable t, when t is a value, the value of s is the largest;

(3) Is there a point M that makes △AQM a right triangle? If it exists, find the coordinates of point m,

If it does not exist, explain why.

Exercise 2 .. (Jiangxi Province) 25. Experiment and inquiry

(1) In figure 1, 2, 3, give the coordinates of the vertices of the parallelogram (as shown in figure), and write the coordinates of the vertices in figure 1, 2, 3, which are,, respectively;

(2) In Figure 4, the coordinates of the vertices of the parallelogram are given (as shown in the figure), and the coordinates of the vertices are obtained (the point coordinates are expressed by the contained algebraic expression);

Induction and discovery

(3) By observing the figure 1, 2, 3 and 4 and exploring the coordinates of the vertices, you will find that no matter where the parallelogram is in the rectangular coordinate system, when the coordinates of its vertices are (as shown in Figure 4), the equivalent relationship between the abscissas of the four vertices is; The equivalent relationship between the ordinate is (without proof);

Application and popularization

(4) There is a parabola and three points in the same rectangular coordinate system. When is the value, there is a point on this parabola, so that a quadrilateral with vertices is a parallelogram? And get all the qualified point coordinates.

Answer:

1. Quadratic function and the shape of quadrilateral

Example 1. Solution: (1) Let y=0 and get or ∴ A (- 1, 0) b (3,0);

Substituting x=2 into the abscissa of point C, the resolution function of ∴C(2, -3)∴ straight line AC is y=-x- 1.

(2) If the abscissa of point P is x(- 1≤x≤2), the coordinates of P and E are respectively: P(x, -x- 1),

E( ∵P is above e, PE=

∴, the maximum value of PE =

(3) There are four such points f, namely

Exercise 1. Solution: (1) From the parabola symmetry axis, the analytical formula can be set as. Substitute the coordinates of a and b into the above formula, and you get

Solve it and get it.

So the analytical formula of parabola is, and the vertex is.

(2)∵ Point is on the parabola, located in the fourth quadrant, and the coordinates are appropriate.

,

∴y<； 0, that is, -y > 0, and -y represents the distance from point E to OA. ∫OA is diagonal,

∴ .

Because the two intersections of the parabola and the axis are (6,0) of (1, 0), and those of the independent variable.

The value range is 1 < < 6.

According to the meaning of the question, when S = 24, that is.

Simplify, get the solution, get it

So there are two points E, namely E 1(3, -4) and E2(4, -4).

Point E 1(3, -4) satisfies OE = AE, so it is a diamond;

Point E2(4, -4) does not satisfy OE = AE, so it is not a diamond.

(2) When OA⊥EF and OA = EF, it is a square, and at this point E.

Coordinates can only be (3, -3).

And the point with coordinate (3, -3) is not on the parabola, so there is no such point e,

Make it square.

Exercise 2. Solution: The function relation set by the coordinates of (1) point is.

Then point on the parabola and find the solution.

The function relation of parabola is (or).

(2) sum is always symmetrical and parallel to the axis.

If the abscissa of a point is 0, its ordinate is 0, that is, when, it is solved. When, it will be solved. When the point moves to or or or,

A quadrilateral whose vertex is a point is a parallelogram.

(3) The point that meets the conditions does not exist. The reasons are as follows: If there is a point that meets the above conditions, then

, (or),

.

You can get it by doing it repeatedly.

, , .

The coordinates of this point are.

However, when ...

There is no such point to form a right triangle that meets the conditions.

Exercise 3. [Solution] (1) Point, point and point's symmetrical points about the origin are, respectively. Let the analytical formula of parabola be

, the solution.

So the analytical formula of parabola is.

(2) The number of points that can be calculated by (1).

If you work too hard, you will lose your foot.

When it's time to exercise.

According to the nature of central symmetry, quadrilateral is parallelogram.

So, so, the area of the quadrilateral. Because it moves until the point coincides with the point, according to the meaning of the question.

Therefore, the relationship is and the value range is.

(3) ,( ).

So there is a maximum.

Tip: You can also use the formula of vertex coordinates.

(4) The quadrilateral can form a rectangle in the process of movement.

According to (2), the quadrilateral is a parallelogram and the diagonal is a rectangle.

So ... So ...

Therefore, get a solution (give up).

So quadrilateral can form a rectangle in the process of movement, at this time.

[Comment] This topic is based on quadratic function, combined with dynamics problem, existence problem and maximum problem. It is a more traditional finale and requires higher ability.

2. Quadratic function and the area of quadrilateral

Example 1. Solution: (1) Solution 1: Suppose,

Substitute any three groups of values of x and y to find the analytical formula.

Let y=0 and find out; Let x=0 and y=-4.

∴ The coordinates of points A, B and C are A (2 2,0), B (-4,0) and C (0 0,4) respectively.

Solution 2: Passing through the point from parabola p (1,-), (-3,),

The equation of the symmetry axis of parabola p is x=- 1,

If the parabola p passes through (2,0) and (-2,4), we can see from the symmetry of the parabola.

The coordinates of points A, B and C are A (2 2,0), B (-4,0) and C (0 0,4) respectively.

(2) From the meaning of the question, and AO=2, OC=4, AD=2-m, so DG=4-2m,

Furthermore, EF=DG, get BE=4-2m, ∴ DE=3m,

∴ =DG？ DE =(4-2m)3m = 12m-6 m2(0 < m < 2)。

Note: It can also be solved by solving Rt△BOC and Rt△AOC, or by establishing the relationship that △BOC is an isosceles right triangle.

(3) When ∵ s defg = 12m-6m2 (0 < m < 2) and ∴m= 1, the area of the rectangle is the largest, and the maximum area is 6.

When the rectangular area is the largest, its vertices are D (1, 0), G (1, 2), F (-2, 2), E (-2, 0),

Let the analytical formula of straight line DF be y=kx+b, and it is easy to know that k=, b=-, Ⅷ.

The analytical expression of parabola p can also be obtained as follows:

Let =, we can find the answer Let the ray DF intersect the parabola p at point n,

Then the abscissa of n is, and the vertical line passing through n with n as the x axis intersects the x axis at h, then there is

= = ,

When the point m is not on the parabola p, that is, when the point m and n do not coincide, the value range of k at this time is

K≠ and k > 0.

Note: If one of the above two conditions is missed by mistake, this step does not score.

If you choose another question:

(2)∵, and AD= 1, AO=2, OC=4, DG=2,

∫, and AB=6, CP=2, OC=4, FG=3,

∴ =DG？ FG=6。

Exercise 1. Solution: Draw a trapezoid OABC. Using the central symmetry characteristic. 1 point

∫A, B, C and M, N, H are symmetrical about the center of point O, respectively.

∴A(0,4),B(6,4),C(8,0)？ 3 points

(Write the coordinates of a wrong point 1 min)

(2) Let the parabola relation of a, b and c be,

∫ parabola intersection A (0 0,4),

Then the parabola relation is .4 points.

Substituting the coordinates of B (6 6,4) and C (8 8,0) into this relation, we get

5AB, the vertical foot is g, then sin ∠ Feg = sin ∠ cab = points.

Solve? 6 points

The parabola relation is: .7 points.

(3)OA = 4，OC=8，∴ AF = 4-m，OE = 8-m.8

∴

OA(AB+OC) AF？ AG OE？ CE's Involve (same as on or about)

(0 < < 4) 10 point

When ∵. ∴, the value of S is the smallest.

And ∵ 0 < m < 4, ∴ there is no value of m, which makes the value of s the smallest. 12 point

(4) When, GB=GF, when, BE = BG. 14 points.

Exercise 3. [Solution] (1) When,,,,

Namely.

(2) When the rubber band just touches the nail,

, , , .

(3) When,

, ,

,

Namely.

Work, to hang your feet.

When,

,

Namely.

or

(4) As shown:

Exercise 4. [Solution] (1) Let the analytical formula of l2 be y=ax2+bx+c(a≠0).

∫ the intersection of l 1 and the x axis is a (-2,0), C (2 2,0), the vertex coordinate is (0,4), and l2 and l1are symmetrical about the x axis.

∴l2 passes through a (-2,0), c (2,0), and the vertex coordinate is (0,4).

∴

∴ a=- 1, b=0, c=4, that is, the analytical formula of l2 is y= -x2+4.

(You can also use vertex, symmetry and other methods to solve)

(2) Let point B (m, n) be any point on L 1: y = x2-4, then n= m2-4 (*).

∵ quadrilateral ABCD is a parallelogram, and point A and point C are symmetrical about the origin O,

∴ B and d are symmetrical about the origin o,

The coordinate of point d is D(-m, -n).

According to formula (*), n=-(m2-4)= -(-m)2+4,

That is, the coordinate of point d satisfies y= -x2+4,

Point D is on l2.

(3) □ABCD can be rectangular.

Let BH⊥x axis pass through point B as H, L 1: Y = X2-4 when passing through point B, and let the coordinates of point B be (x0, x02-4).

Then OH=| x0|, BH=| x02-4|.

It is easy to know that □ABCD is rectangular if and only if BO= AO=2.

In Rt△OBH, from Pythagorean theorem, | x0|2+| x02-4|2=22,

(x02-4)( x02-3)=0, ∴ x0 = 2 (excluding), x0= 3.

Therefore, when the coordinate of point B is B(3,-1) or B'(-3,-1), □ABCD is a rectangle. At this time, the coordinates of point D are D(-3, 1) and D'(3, 1) respectively.

Therefore, there are only two qualified rectangles, namely rectangle ABCD and rectangle AB'CD'.

Let the straight line AB intersect the Y axis at E, obviously, △AOE∽△AHB.

∴ io =, ∴

∴ EO=4-2。

According to the symmetry of the graph, the overlapping part of rectangle ABCD and rectangle AB'CD is an area of

s = 2sδACE = 2× 12×AC×EO = 2× 12×4×(4-23)= 16-83。

3. Dynamic exploration of quadratic function and quadrilateral.

Example 1. Solution: (1) If ∠APD and ∠OPF are equally divided by known PB, PD and PF overlap, ∠ BPE = 90. ∴∠ OPE+∠ APB = 97。

∴Rt△POE∽Rt△BPA.

That is. ∴ y = (0 < x < 4)。

And when x=2, y has the maximum value.

(2) It is known that both △PAB and △POE are isosceles triangles, and P (1, 0), E (0, 1) and B (4 4,3) can be obtained.

Let the parabola passing through these three points be y = AX2+BX+C, then ⅷ

y=。

(3) From (2), it can be known that ∠ EPB = 90, that is, the conditions are met when point Q and point B coincide.

The straight line PB is y = x- 1, and intersects the y axis at the point (0,-1).

Translate PB upward by 2 units through point E (0, 1).

The straight line is y = x+ 1.

From ∴ q (5 5,6).

So there are two points Q (4 4,3) and (5,6) on this parabola that satisfy the conditions.

Example 2. Solution: (1) Solve the equation x2- 10x+ 16 = 0 to get x 1 = 2, x2 = 8 .....................................1min.

∵ point b is on the positive semi-axis of x axis, point c is on the positive semi-axis of y axis, and ob < oc.

∴ The coordinates of point B are (2,0), and the coordinates of point C are (0,8).

The symmetry axis of parabola Y = AX2+BX+C is straight line X =-2.

According to the symmetry of parabola, the coordinate of point A is (-6,0) ..................... 4 points.

(2)∵ point c (0 0,8) is on the image of parabola y = ax2+bx+c.

∴ c = 8. Substitute A (-6,0) and b (2 2,0) into the expression to obtain

solve

The expression of parabola is y = X2x+8. .........................................................................................................................................................

(3) According to the meaning of the question, AE = m, then BE = 8-m,

∵OA=6,OC=8,∴AC= 10

∫ef∑ AC ∴△BEF∽△BAC

namely

∴EF=

∴ =∴FG=？ =8 meters

∴ s = s △ BCE-s △ BFE = (8m) × 8-(8m) (8m)

= (8-m) (8-8+m) = (8-m) m =-m2+4m ...................10.

The range of independent variable m is 0 < m < 8. ...............................................................................................................................................................

(4) existence.

Reason: ∫s =-m2+4m =-(m-4)2+8 and -< 0,

When m = 4, s has a maximum value, and the maximum value of s = 8. ...............................................................................................................................................

∵ m = 4, and the coordinate of ∴ point E is (-2,0).

∴△BCE is an isosceles triangle ........................ 14 minute.

(The above answers are for reference only. If there are other methods, please refer to the grading. )

Example 3 solution: (1) equals to

The reason is that quadrilateral ABCD and EFGH are both rectangles,

therefore

That is:

(2) AB = 3, BC = 4, AC = 5, let AE = X, then EC = 5-X,

So, that is

The formula is:, so when,

The maximum value of s is 3.

(3) When AE = AB = 3 or AE = BE = or AE = 3.6, it is an isosceles triangle.

Exercise 1. Solution: After (1) point M 1 minute (2)t seconds,

So, ∫ = ∴ ∴

∴ ∴

∵∴ When appropriate, the value of S is the largest.

(3) existence. Let NB=t, OM=2t, ∴= = t seconds later.

① If yes, it is the height on the bottom edge of isosceles RTδ.

∴ is the center line of the bottom edge.

∴ The coordinate of this point is (1, 0).

② If, at this time, it overlaps with ∴∴∴.

∴ The coordinate of this point is (2,0)

Exercise 2. Solution: (1),.

(2) The vertical lines passing through each point are taken as the axis, and the vertical feet are respectively,

They are over-performing, on the point.

In the parallelogram,

.

.

Say it again,

.

, .

Settings. From, get.

By, by ..

(3), or,.

(4) If it is the diagonal of a parallelogram, it can be obtained from (3). In order to make it on a parabola,

Yes, it is.

At this time.

If it is the diagonal of a parallelogram, it can be obtained from (3), and it can also be obtained.

If it is the diagonal of a parallelogram, it can be obtained from (3), and it can also be obtained.

To sum up, when there are points on a parabola, a quadrilateral with vertices is a parallelogram.

Qualified points are,,.

Exercise 3. Solution: (1) from Rt△AOB≌Rt△CDA.

OD=2+ 1=3，CD= 1

The coordinate of point ∴c is (-3, 1).

∵ The parabola passes through point C,

∴ 1= (-3)2 a+(-3)a-2,∴.

The analytical formula of parabola is.

(2) There are points P and Q (on the right side of the symmetry axis) on the parabola, so that the quadrilateral ABPQ is a square.

Make a square ABPQ with AB side on the right side of AB. P is PE⊥OB in E, QG⊥x axis in G,

It can be proved that △ △PBE?△AQG?△ abalone,

∴PE=AG=BO=2,BE=QG=AO= 1，

∴∴ The coordinates of point P are (2 1), and the coordinates of point Q are (1,-1).

Press (1) parabola.

When x = 2, y = 1, when x =, 1, y =- 1.

P and q are on a parabola.

So there are points P(2, 1) and Q (1,-1) on the parabola (the right side of the symmetry axis), which makes the quadrilateral ABPQ a square.

⑵ Another solution: There are points P and Q (on the right side of the symmetry axis) on the parabola, so that the quadrilateral ABPQ is a square.

Extend the intersection parabola of CA to q, let b be BP∑CA to p, and connect PQ. Let the analytical expressions of straight lines Ca and BP be y=k 1x+b 1 and y=k2x+b2, respectively.

∫A(- 1，0)，C(-3， 1)，

∴ The analytical formula of ∴CA's, similarly the analytical formula of BP is,

The coordinates of Q point for solving the equations are (1,-1), and the coordinates of P point are (2 1).

According to Pythagorean theorem, AQ = BP = AB =, and ∠ Baq = 90.

∴ quadrilateral ABPQ is a square. So there are points P(2, 1) and Q (1,-1) on the parabola (the right side of the symmetry axis), which makes the quadrilateral ABPQ a square.

⑵ Another solution: There are points P and Q (on the right side of the symmetry axis) on the parabola, so that the quadrilateral ABPQ is a square.

As shown in the figure, the line segment CA translates to AQ in the direction of CA,

The corresponding point of ∵ C (-3, 1) is A (- 1, 0), and the corresponding point of ∴ A (- 1, 0) is Q (1,-1).

∫∠BAC = 90，AB=AC

∴ quadrilateral ABPQ is a square. It is proved that P(2, 1) and Q (1,-1) are all on parabola.

Conclusion (2) holds,

The proof is as follows: Even if EF, if F is the extension line of FM∨BG∨AB, then △ AMF ∽△ AB.

∴ 。 From (1), it can be known that △ABC is an isosceles right triangle,

∴∠ 1=∠2=45 。 ∵AF=AE,∴∠AEF=∠ 1=45 .∴∠ EAF = 90，ef⊙o？ The diameter of.

∴∠EBF=90 .∵FM∥BG,∴∠MFB=∠EBF=90，∠M=∠2=45，

∴BF=MF，

∴