As shown in the figure, in ladder ABCD, ab ‖ CD, e is the midpoint of DC, the line be intersects AC at F, and the extension line of ad intersects g; verification: EF · BG = BF · eg

As shown in the figure, in ladder ABCD, ab ‖ CD, e is the midpoint of DC, the line be intersects AC at F, and the extension line of ad intersects g; verification: EF · BG = BF · eg

It is proved that: ∵ ab ∥ CD, ∥ GDE = ∥ gab, ∥ GED = ∥ GBA, ∥ CEF = ∥ ABF, ∥ ECF = ∥ BAF. ∥ CEF ∥ ABF, ∥ DGE ∥ AGB. ∥ EF: BF = EC: AB, eg: BG = de: ab. ∥ de = EC, ∥ EF: BF = eg: BG. ∥ EF · BG = BF · eg

As shown in the figure, in rectangle ABCD, ab = 15cm, point E is on ad, and AE = 9cm, connect EC, fold rectangle ABCD along straight line be, and point a just falls on point a 'on EC, then a' C=______ cm．

The rectangle, the rectangle, the rectangle, the rectangle, the rectangle, the rectangle, the rectangle, the rectangle, the ab = CD = 15cm, the \\\\\\\\\\\\\\\\\\\ = 15cm, \\\\\\\\\\ = 15cm = CD = 15cm, \\\\\\\\\\\\\\\\\\\xcm, then BC = ad =De + AE = x + 9 (CM), in RT △ a ′ BC, BC2 = a ′ B2 + a ′ C2, i.e. (x + 9) 2 = x2 + 152, the solution is: x = 8, 〈 a ′ C = 8cm

The area of rectangle ABCD is 24 square centimeters, DF is equal to 1:2, be is equal to EC, what is the ratio of the area of shadow part to the area of rectangle
F is between DC, e is between BC, and the shadow is after. The answer is A1: 2, B3: 4, C5: 12, D7: 12

Let DC = 3A, BC = 2B
Sabcd=24=6ab
S non shadow = AB + 3AB / 2 = 5ab / 2
Then s shadow = sabcd-s non shadow = 7ab / 2
What is the ratio of the area of the shadow part to the area of the rectangle? 7:12
D

As shown in Figure 1, in the square ABCD, e is a moving point on CD, connecting the intersection diagonal BD of AE to point F, passing through point F to make FG ⊥ AE intersection BC to point G
1. Proving AF = FG
2. Connect eg, when BG = 3, de = 2, find the length of eg

1. Connect FC, because ad = CD & nbsp; & nbsp; DF = DF & nbsp; & nbsp; & nbsp; ∠ ADF = CDF
∴△ADF≅△CDF
∴AF=CF
∠DAF=∠DCF
The remaining angles of equal angles are equal
Because ∠ ABG = ∠ AFG = RT ∠
∴∠ABG+∠AFG=180°
∴∠FAB+∠FGB=180°
■ ∠ FGC = ∠ FAB (the same complement of ∠ FGB)
∴∠FGC=∠FCG
∴AF=FG
Note: it will be very convenient to use four points to prove
2. Even AG, △ AFG are isosceles right triangles,
∴∠FAG=45°
∴∠DAE+∠BAG=45°
Rotate △ ade clockwise around point a to the position of △ ABH,
Ah = AE & nbsp; & nbsp; & nbsp; Ag = Ag & nbsp; & nbsp; & nbsp; de = BH
∠HAG=∠DAE
∴∠HAB+∠BAG=∠DAE+∠BAG=45°=∠EAG
△BAG≅△EAG
∴EG=HG=HB+BG=DE+BG
So eg = 3 + 2 = 5

It is known that the area of square ABCD is 256, point E is on the extension line of AB, point F is on ad, and EC is perpendicular to FC. The area of triangle CEF is 200

Because ∠ FCD + ∠ BCF = 90 degree ∠ ECB + ∠ BCF = 90 degree, so ∠ FCD = ∠ ECB, because ABCD is a square, so CD = BC, ∠ CDF = ∠ CBE, so △ CDF ≌ △ CBE, so CF = Ce (1 / 2) * ce ^ 2 = 200, the solution is CE = 20, because the area of square ABCD is equal to 256, so BC = 16be ^ 2 = CE ^ 2-bc ^ 2be ^ 2 = 400-256b

As shown in the figure, in the square ABCD, ab = 4A, e is the midpoint of AB, DF = 3af (1) find the length of EF (2) prove: △ CEF is a right triangle
The first question has been proved. The second question is how to write it?

Proof: by Pythagorean theorem,
EF^2=AF^2+AE^2=a^2+(2a)^2=5a^2,
FC^2=DF^2+CD^2=(3a)^2+(4a)^2=25a^2,
EC^2=BE^2+BC^2=(2a)^2+(4a)^2=20a^2,
EF^2+EC^2=FC^2
A right triangle is a right triangle

In the isosceles trapezoid ABCD, AB is parallel to CD, ∠ a = 60 °, DB bisects ∠ ABC, and the trapezoid perimeter is 30cm
. in a minute

Because AB is parallel to CD, so ﹥ a + ﹥ ADC = 180 °, that is, ﹥ ADC = 120 ° and quadrilateral ABCD is isosceles trapezoid, so ﹥ ABC = 60 ° and ad = CB. Because DB bisects ﹥ ABC, so ﹥ abd = 30 ° and ﹥ BDC = 30 °, so in triangle abd, ﹥ abd = 90 ° and triangle abd is right triangle

As shown in the figure, if ad ‖ BC, ab = CD, ad = 2, BC = 6, ∠ B = 60 ° in trapezoidal ABCD, the perimeter of trapezoidal ABCD is______ ．

Let a pass through point a as AE ⊥ BC at point E, ∵ ABCD is trapezoid, ab = CD, ∵ quadrilateral ABCD is isosceles trapezoid, ∵ be = 12 (BC-AD) = 2, in RT △ Abe, ab = becos ∠ Abe = 4, so the circumference of trapezoid ABCD = AD + AB + BC + DC = 16

As shown in the figure, in the trapezoidal ABCD, ad ‖ BC, ab ‖ De, AF ‖ DC, e and F are on the side BC, and the quadrilateral aefd is a parallelogram. (1) what is the equivalent relationship between AD and BC, please explain the reason; (2) when AB = DC, prove that the parallelogram aefd is a rectangle

(1) Ad = 13bc. The reasons are as follows: ∵ ad ∥ BC, ab ∥ De, AF ∥ DC, ∵ quadrilateral abed and quadrilateral AFCD are parallelograms. ∵ ad = be, ad = FC, and ∵ quadrilateral aefd are parallelograms, ∵ ad = EF. ∵ ad = be = EF = FC. ∵ ad = 13bc

In the quadrilateral ABCD, if a (0,0), B (1,0), C (2,1), D (0,3) rotate around the y-axis for one circle, the volume of the rotator is______ ．

If the quadrilateral ABCD rotates around the y-axis for one circle, the upper part of the rotator is a cone, and the lower part is an inverted frustum. Therefore, V-cone = 13 π r2h = 13 π × 22 × 2 = 83 π, V frustum = 13 π H (R2 + R2 + RR) = 13 π × 1 × (22 + 12 + 2 × 1) = 73 π, v = V-cone + V frustum = 5 π