Two elementary geometry problems 1. In rectangle ABCD, point O is the midpoint of AC, AC = 2Ab, extend AB to g, make BG = AB, connect go to intersection BC to e, extend go to intersection ad to point F, prove: quadrilateral aecf is diamond 2. Make three equilateral triangles △ abd, △ BCE, △ ACF on the same side of BC with △ ABC three times (1) Quadrilateral what is ADEF and why? (2) When △ ABC satisfies what condition, the quadrilateral ADEF is a rectangle (3) When △ ABC satisfies what conditions, the quadrilateral ADEF is a diamond, why? (4) When △ ABC satisfies what conditions, quadrilateral with a, D, e, F as vertex does not exist? Thank you~

Two elementary geometry problems 1. In rectangle ABCD, point O is the midpoint of AC, AC = 2Ab, extend AB to g, make BG = AB, connect go to intersection BC to e, extend go to intersection ad to point F, prove: quadrilateral aecf is diamond 2. Make three equilateral triangles △ abd, △ BCE, △ ACF on the same side of BC with △ ABC three times (1) Quadrilateral what is ADEF and why? (2) When △ ABC satisfies what condition, the quadrilateral ADEF is a rectangle (3) When △ ABC satisfies what conditions, the quadrilateral ADEF is a diamond, why? (4) When △ ABC satisfies what conditions, quadrilateral with a, D, e, F as vertex does not exist? Thank you~

1. Prove that triangle AOF is equal to triangle Coe

Geometry problems (2)
1. Given the extension line AB to C, let BC = 2Ab, take the midpoint D of AC, if BC = 6cm, find the length of BD (because so)
2. From 2:11 to 2:18, what's the rotation angle of the minute hand of the clock? How many minutes does the minute hand of the clock coincide with the hour hand for the first time from 4:00

1. Let AC = x, BD = (1 / 2) x, CD = (3 / 2) x, because BD + CD = 6cm, so x = 3cm, BD = 1.5cm
2. Divide 360 degrees equally into 60 parts, then the minute hand turns 6 degrees in 1 minute, after 7 minutes from 11 minutes to 18 minutes, the rotation angle is 7 * 6 = 42 degrees
Divide 30 degrees equally into 60 parts, then turn 0.5 degrees clockwise for 1 minute, and set the overlap after X minutes
When the minute hand goes X minutes, it turns 6x degrees, the hour hand turns 0.5x degrees, and the difference between the hour hand and the minute hand is 120 degrees at 4 o'clock
6X = 0.5x + 120, x = 240 / 11

Two geometric problems
The shapes of the following objects are cylinders ()
A. An unused pencil with round upper and lower sides
B. Fully inflated bicycle inner tube
A pentaprism has () vertices, () edges, and how many faces?

First question a
Question 2: 10 vertices, five ridges and three faces (upper and lower sides)

The π of two problems is about 3.14
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First question:
Second question:
A 18.84 cm long rectangular paperboard is made into a cylinder. The surface area of the cylinder is 175.72 square cm. What is the width of the rectangular paperboard?

I thought about it for five minutes and finally came up with it
Divide 6 * 6 * 3.14 by 4 = 28.26 to calculate the total area
3 * 3 = 9 work out the right side of the rectangle below
Divide 3 * 3 * 3.14 by 4 = 2.355 to calculate the quarter circle in the block on the right of the rectangle below
9-2.355 = 6.645 area except quarter circle
9 + 6.645 = 15.645 blank area
28.26-15.645 = 12.615
I don't know, right
Second question;
18.84/3.14=6
6/2=3
3*3*3.14*2=56.52
175.72-56.52=119.2
119.2 / 18.84 = ~ not enough··············
Keep integers
It's six
Oh, in addition, I take the length as the circumference

Two high school mathematics problems, please help solve, is about geometry, to detailed process and accurate steps, thank you!
1. The straight line B ∥ C is known, and the straight line a intersects with B and C. It is proved that the straight lines a, B and C are coplanar
2. Verification: two straight lines with different planes cannot be perpendicular to one plane at the same time
I hope you can help me solve these two problems. My geometry is really bad. Thank you. I will add points to you according to the situation. Thank you!

1. The intersection of line a and B, C is m, N, BC determines plane β, then m, n are in plane β, m, n are on line a, so line ABC is coplanar
2. Suppose that the out of plane lines a and B are perpendicular to the plane β at the same time. According to the fact that two lines perpendicular to the same plane are parallel, then a / / B, a and B are coplanar and out of plane

It is known that the area of triangle ABC is s, the acute angle between plane ABC and plane a is V, and the orthographic projection of triangle ABC in plane a is triangle a'b'c ', and its area is s'. Prove: s' = scosv

Given any △ ABC, D is the interior point of △ ABC, e is the exterior point of △ ABC, connecting ad, BD, be, EC, ed. if ∠ abd = ∠ CBE, ∠ bad = ∠ BCE, prove: △ ABC ∽ DBE

Because ∠ abd = ∠ CBE, ∠ bad = ∠ BCE
So △ abd ∽ CBE
So AB / BD = CB / be
And because ∠ abd = ∠ CBE
Therefore, ABC = abd + DBC = CBE + DBC = DBE
So △ ABC ∽ DBE (triangles with proportional sides and equal angles are similar)
The key to solve the problem is to transform AB, BD, CB and be between different triangles

Two cylinders, a cylinder height: 107cm, width: 108CM, B cylinder height: 116Cm, width: 120cm
Put a in cylinder B, find the volume between two cylinders AB, and then convert the volume into square meters
Wrong number. It's a cube

Pi * (1.2 / 2) (1.2 / 2) * 1.16-pi * (1.08 / 2) (1.08 / 2) * 1.07 = 0.3317 M3

（1） It is known that ab = AC, angle a = 90 °, P is the midpoint of BC, PE = PD, and be = ad
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（2） Square ABCD, e is a point on CD, angle ECP = 135 °, AE = EP, prove AE perpendicular to EP
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What I want to remind you is: what you need to think about is you, think about it, the first question is not the one you saw in the past, it has been adapted! And the second question, think about it carefully, read it well!

The first problem is that PG and PF are parallel to AB and AC respectively through point P. PGD and PFE are congruent, so EF = DG, and because PG and PF are median lines, ab = AC, so CG = AF, so be = ad
Question 2 Wait a minute, add points first

A geometric problem about circle in the third grade of junior high school
It is known that the radius of 70 o is 5, the distance OA from point a to the center O is 3, and the shortest chord length of point a is obtained

Let a be EF ⊥ OA, and let o be e, F, EF
((1/2)EF)^2+OA^2=OE^2=OF^2
(1/4)EF^2+3^2=5^2
EF=8
Prove the shortest ef
Let there be a chord cdoh passing through point a
(1/2)EF=√5^2-OA^2
(1/2)CD=√5^2-OH^2
EF > CD and its contradiction