It is known that e is the middle point on the side BC of square ABCD, f is a point on CD, AE bisects ∠ BAF

Method 1: to make em \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\3. Em =AM=12AF∵EM=12（AB+CF），∴AF=AB+CF=BC+CF．

It is known that, as shown in the figure, in square ABCD, e is the midpoint of BC, and point F is on CD, AF = BC + CF, and ∠ FAE = ∠ Bae is proved

Because point E is the midpoint of BC, OB = CF has BC + CF = AB + Bo = Ao, so AF = Ao △ AOF is isosceles triangle and E is the midpoint of of, so ∠ OAE = ∠ EAF is ∠ BAE = ∠ FAE

In square ABCD, e is the midpoint of BC, f is on CD, and the angle FAE = angle BAE

The verification should be: AF = BC + CF
When the extension line of extended DC and AE intersects g, because AB is parallel to DC, AFG is isosceles triangle (angle fag = angle bag = angle FGA), AF = FG; triangle ECG is equal to triangle Abe (opposite vertex angle is equal, internal stagger angle is equal, right angle is equal, e is midpoint), CG = AB = BC, AF = FG = FC + CG = FC + BC)

It is known that the quadrilateral aefd and the quadrilateral ebcf are parallelograms as shown in the figure

It is proved that both the ∵ quadrilateral aefd and the quadrilateral ebcf are parallelograms, and the ≌ quadrilateral ABCD is parallelograms. AB = CD. In △ Abe and △ DCF, AE = dfbe = CFAB = CD. Abe ≌ DCF

Verification: in the parallelogram ABCD, ac2 + BD2 = AB2 + BC2 + Cd2 + DA2

It is proved that if we make the extension line of de ⊥ BA at point E and CF ⊥ AB intersecting AB at point F, then ⊥ AED = ⊥ BFC = 90 °. ∵ quadrilateral ABCD is a parallelogram, ⊥ AB = DC, ab ∥ CD, ⊥ DAE = ⊥ CBF, in △ ade and △ BCF, ⊥ DEA = ⊥ CFB ≌ DAE = ⊥ cbfad = BC, ≌ ade ≌ BCF (AAS), ≌ AE = BF, de = cf. in RT △ DBE and RT △ CAF, ac2 = af2 + CF2 = CF2 + (AB + a) E) Then ac2 + BD2 = CF2 + AB2 + AE2 + 2Ab · AE + CF2 + ab2-2ab · AE + AE2 = (CF2 + AE2) + (CF2 + AE2) + AB2 + AB2 = AB2 + BC2 + Cd2 + DA2

The parallelogram ABCD is known and verified: AC & # 178; + BD & # 178; = AB & # 178; + BC & # 178; + CD & # 178; + Da & # 178;
AC BD is the opposite side of parallelogram

We can get this conclusion by phasor method. Vector AC = vector AB + vector ad, vector BD = vector AB + vector BC
Square, square sum can get the above conclusion

In the parallelogram ABCD, we prove: | ab | & # 178; + | BC | & # 178; + | CD | & # 178; + | Da | & # 178; = | AC | & # 178; + | BD|

If you can use vector knowledge, it is very simple. AC 124; AC | ^ 2 + | BD | ^ 2 = | AB + BC ^ 2 + | ab-ad | ^ 2 = | AB + BC | AB + BC | ^ 2 + | AB + BC | ^ 2 + | AB + BC (AB + BC) (AB + BC) + (ab-bc) (ab-bc) (ab-bc) (ab-b-bc) is very simple. AC | AC | AC | ^ 2 + | AC | ^ 2 + | AC ^ 2 + | ^ 2 + | ^ ^ ^ ^ ^ ^ 2 + | ^ ^ ^ ^ ^ ^ 2 + | AC ^ 2 + | ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ it's all

In the rectangular coordinate system, O is the origin, a's coordinate is (2,0), B's coordinate is (0,2), connecting AB and C on the x-axis, and a, B and C form a right triangle,
Try to find all the C points that meet the conditions, and write the coordinates

∵ the coordinates of a and B are (2,0) and (0,2), connecting AB and C on the x-axis
There are two C points satisfying the conditions: when ∠ BCA = 90 ° C (0,0); when ∠ ABC = 90 ° C (- 2,0)

In the rectangular oabc plane rectangular coordinate system (o is the coordinate origin), point a is on the X axis, and point C is on the Y axis
As shown in the figure, the rectangular oabc is in the plane rectangular coordinate system (o is the coordinate origin), point a is on the x-axis, point C is on the y-axis, and the coordinates of point B are (- 2,2 times the root 3), point E is the midpoint of BC, point H is on OA, and ah = 1 / 2. Hg and EB passing through point h and parallel to the y-axis intersect at point g. now fold the rectangle so that vertex C falls on Hg and coincides with point D on Hg, the crease is EF, and point F is the intersection of crease and y-axis
(1) Find the degree of ∠ CEF and the coordinates of point d (2) find the function expression of the line where the crease EF is located (3) if point P is on the line EF, when △ PFD is an isosceles triangle, how many points P satisfy the condition? Ask for the coordinates of point P,

(1) because the OA and OC of the rectangular oabc coincide with the coordinate axis, the coordinate of point B is (- 2,2 * 3 ^ (1 / 2))
Then OA = BC = 2, OC = AB = 2 * 3 ^ (1 / 2)
E is the midpoint of BC, then CE = be = BC / 2 = 1
AH=1/2
Hg ‖ Y axis, then BG = ah,
EG=BE-BG=1-1/2=1/2
Because △ DEF is folded from RT △ CEF, so △ DEF is RT △ and
△DEF≌△CEF
∠DEF=∠CEF,DE=CE=1
In RT △ DEG, cos ∠ DEG = eg / de = (1 / 2) / 1 = 1 / 2, then ∠ DEG = 60 degree
And ∠ def = ∠ CEF,
∴∠CEF=(180°-∠DEG)/2=(180°-60°)/2=60°
The abscissa of point D is: 1 / 2-2 = - 3 / 2
The ordinate is: 2 * 3 ^ (1 / 2) - 1
That is, its coordinates are (- 3 / 2,2 * 3 ^ (1 / 2) - 1)
(2) it is known from (1) that ∠ CEF = 60 °, that is, the angle between crease EF and x-axis is 180 ° - 60 ° = 120 °
Let the function of the line be y = KX + B
Then k = Tan 120 ° = - 3 ^ (1 / 2)
Because the straight line passes the e point (- 1,2 * 3 ^ (1 / 2))
Substituting into the linear equation, 2 * 3 ^ (1 / 2) = - 3 ^ (1 / 2) * (- 1) + B, B = 3 ^ (1 / 2)
So the function of straight line EF is: y = - 3 ^ (1 / 2) x + 3 ^ (1 / 2)
(3) it can be seen from the above that the function of straight line EF is: y = - 3 ^ (1 / 2) x + 3 ^ (1 / 2)
Then, the coordinates of point F are: (0,3 ^ (1 / 2))
And ∠ DFP = 30 °, DF = CF = 2 * 3 ^ (1 / 2) - 3 ^ (1 / 2) = 3 ^ (1 / 2)
If point P is on EF, there are three cases to make △ PFD an isosceles triangle
①P1D=P1F
If p1q ⊥ DF is over P1, then q is the midpoint of DF
If FQ = DF / 2 = (3 ^ (1 / 2)) / 2, then p1f = QF / cos ∠ dfp1 = 1
So the abscissa of point P1 is: - p1f * cos60 ° = - 1 / 2
The ordinate is: 3 ^ (1 / 2) + p1f * sin60 ° = (3 * 3 ^ (1 / 2)) / 2
That is, P1 is (- 1 / 2, (3 * 3 ^ (1 / 2)) / 2)
②DF=P2F
At this point, p2f = 3 ^ (1 / 2)
So the abscissa of point P2 is: - p2f * cos60 ° = - (3 ^ (1 / 2)) / 2
The ordinate is: 3 ^ (1 / 2) + p2f * sin60 ° = (5 * 3 ^ (1 / 2)) / 2
That is, P3 is (- (3 ^ (1 / 2)) / 2, (5 * 3 ^ (1 / 2)) / 2)
③DF=DP3
If Dr ⊥ p3f passes through point D, then R is the midpoint of p3f
In RT △ DFR, FR = DF / cos ∠ dfp3 = 2, then FP3 = 4
So the abscissa of point P3 is: - p3f * cos60 ° = - 2
Ordinate: 3 ^ (1 / 2) + p3f * sin60 ° = 3 * 3 ^ (1 / 2)
That is, P3 is (- 2,3 * 3 ^ (1 / 2))
Oh, it's written about me for a long time

As shown in the figure, the rectangular oabc is in the plane rectangular coordinate system (o is the coordinate origin), point a is on the X axis, point C is on the Y axis, and the coordinate of point B is
They are (- 2,2 times radical 3), point E is the midpoint of BC, point h is on OA, and ah = 1 / 2, Hg and EB passing through point h and parallel to y axis intersect at point g. now fold the rectangle so that vertex C falls on Hg and coincides with point D on Hg, crease is EF, and point F is the intersection of crease and Y axis
Q: (1) find the degree of ∠ CEF and the coordinate of point D
(2) Find the function expression of the straight line where the crease EF is
(3) If point P is on the straight line EF, when △ PFD is an isosceles triangle, how many points P satisfy the condition? How many coordinates of point P are requested,

It is known that e is the middle point on the side BC of square ABCD, f is a point on CD, AE bisects ∠ BAF