The side length of square ABCD is 4, M is the middle point of AD, the moving point e moves on the line AB, connects EM and extends the intersection line CD and F, and makes the intersection line of EF through M BC and G, connect eg and FG, prove that let AE = x, the area of △ EGF = y, find the analytic formula and value range of Y and X. can △ GEF be an equilateral triangle during the movement of 2 points e? Please explain the reason

2011 Kingsoft twenty-fifth the second mock exam

In the square ABCD, M is the middle point of AD, the moving point E is on the line AB, connecting EM and extending ray CD, intersecting F, passing m as the vertical line of EF, intersecting BC and connecting eg
In the square ABCD, M is the middle point of AD, the moving point E is on the line AB, connecting EM and extending the ray CD, intersecting F, passing m as the vertical line of EF, intersecting BC with G, connecting eg and FG. Find the value of me / mg

Mn ⊥ BC in N, EP ⊥ CD in P
Then ∠ EMN = ∠ MGN (both are the remainder of ∠ NMG)
And ∠ EMN = ∠ EFP
∴∠EFP=∠MGN
And EP = Mn
∴Rt△EFP≌Rt△MGN
∴EF=MG
∵ m is the midpoint of AD
Yizheng me = MF
∴ME/MG=1/2

In trapezoidal ABCD, ad ∥ BC and E are the upper moving points of BC. From e to the two waists AB and CD, the vertical lines EF and em are made. The vertical feet are f and m, and the vertical lines CG of AB are made through C
And CG = EF + em
It is proved that the trapezoid ABCD is isosceles trapezoid
emergency

It is proved that: extend me to h, so that EH = EF. Connect ch, then: CG = HM, because HM is vertical to AB, CG is vertical to AB, so hm / / CG, so HMGC is rectangular, then angle H = 90 degree, angle HEC = angle MEB (opposite vertex angle), angle H = angle EMB = 90 degree, so angle B = angle ECH.EF=EH EC = EC, EFC and EHC are right triangles, so the two triangles are congruent

As shown in the figure, we know that △ ABC is a square ABCD and ACGF, and M is the midpoint of BC

Extend am to h, make MH = am, connect HC
Because MH = am, BM = MC, angle BMA = angle CMH
So the triangle BMA is equal to the triangle CMH
So HC = Ba, angle ABC = angle BCH
So HC / / BA
Because in square ABCD and ACGF, the angle EAB = angle fac = 90 degrees
So EAF + BAC = 360-90-90 = 180 degrees
Because the angle ABC + angle BCA + angle BAC of triangle ABC is 180 degrees
So angle EAF = angle ABC + angle BCA
Because angle ABC = angle BCH
So EAF = BCH + BCA = HCA
Because EA = Ba, AF = AC in square ABCD and ACGF
Because HC = ba
So EA = HC, AF = AC
Because angle EAF = angle HCA
So the triangle EAF is equal to the triangle HCA
So EF = ah
Because MH = am = 1 / 2ah
So am = 1 / 2ef

It is known that E and F are the points on AD and CB on the opposite side of () ABCD respectively, and de = BF, EM ⊥ AC in M, FN ⊥ AC in N, EF intersects AC in o
The bracket in front of ABCD is a space on the test paper. I don't know if it is less. If it is less, it should be (parallelogram)

Because ABCD is a parallelogram, so ad = BC, angle EAO = angle FCO, because de = BF, so AE = ad-de = bc-bf = CF, because EAO = angle FCO, angle AOE = angle COF, AE = CF, so triangle AOE is equal to triangle COF, so EO = fo, because EM is perpendicular to AC, FN is perpendicular to AC, so angle

There are four colors in the rectangle, green and yellow account for 50% of the total area, green area accounts for 15% of the total area, yellow is 12 square meters more than green, seeking green surface

There are four colors in the rectangle. Green and yellow account for 50% of the total area. Green area accounts for 15% of the total area. Yellow is 12 square meters more than green?
What percentage of the total area is yellow
50％－15％＝35％
How many percent more yellow area than green area
35％－15％＝20％
Rectangular area:
12 △ 20% = 60 (M2)
Green area:
60 × 15% = 9 (M2)
Answer: green area is 9 square meters
This is an arithmetic method. The equation is as follows:
Setting: the total area is x kilometers
（50％x－15％x）－15％x＝12
20％x＝12
x＝60
60 × 15% = 9 (M2)
Answer: green area is 9 square meters

The rectangle is divided into four different triangles, the area of yellow triangle is 28cm, the area of red triangle is 12cm, what is the area of blue triangle?
The area of green triangles accounts for 15% of the area of the rectangle. The sum of the heights of yellow and green triangles is equal to the width of the rectangle. The sum of the heights of red and blue triangles is equal to the length of the rectangle,
Each side of a rectangle is an edge of a triangle

28
Yellow + Green = Red + blue = half area of rectangle
Green 15%, so yellow 35%
Rectangle area = 28 △ 35% = 80
So the area of blue = 80 △ 2-12 = 28

In trapezoidal ABCD, CD / / EF / / AB, CD = 2, ab = 4, EF bisects the area of trapezoidal ABCD and calculates the length of EF

Let DH ⊥ AB be h, and let EF be g
The area of trapezoidal affe is equal to that of efcd
(CD + EF) * dg / 2 = (EF + AB) * GH / 2 & nbsp; & nbsp; (2 + EF) * DG = (EF + 4) * GH
also
DG: (ef-cd) = DH: (ab-cd) = GH: (ab-ef) & nbsp; & nbsp; yes & nbsp; & nbsp; DG * (4-ef) = GH * (EF-2)
therefore
(EF+2)*(EF-2)=(4+EF)*(4-EF)
EF = √ 10 ≈ 3.16 & nbsp; & nbsp; & nbsp; that is, the length of EF is the root 10, which is about 3.16

In the trapezoidal ABCD, e is on ad, f is on BC, CD / / EF / / AB, CD = 2. AB = 4. EF divide the area of the trapezoidal ABCD into two equal parts and find the EF length

Let the height of trapezoidal aefb be H1, the height of trapezoidal cdef be H2, and the height of trapezoidal ABCD be H
S trapezoid cdef = (CD + EF) H2 / 2 = trapezoid ABCD / 2 = (AB + CD) H / 4  h2 = 3H / (EF + 2) (1)
S trapezoid aefb = (AB + EF) H1 / 2 = trapezoid ABCD / 2 = (AB + CD) H / 4  H1 = 3H / (EF + 4) (2) (1) + (2) then H1 + H2 = 3H / (EF + 2) + 3H / (EF + 4), that is, H = 3H / (EF + 2) + 3H / (EF + 4)
Divide by H to get 1 = 3 / (EF + 2) + 3 / (EF + 4) EF = root 10

CD is parallel to Fe, AB is parallel to AB, CD = 2, ab = 4, EF bisects the area of trapezoid ABCD, and finds the length of Fe

(CD + EF) H1 = 1 / 2 * (CD + AB) H (H1: EF, the height of the trapezoid formed by CD, H: the height of the original trapezoid) = > (2 + EF) H1 = 3H = > H1 / h = 3 / (2 + EF) make DG parallel to BC intersection through D, EF, AB parallel to g, H = > dcbh is parallelogram, BH = CD = 2 = > ah = 4-2 = 2EG = ef-cd = ef-2h1, h is the height of triangle DEG, Dah respectively