In the quadrilateral ABCD, point O is the midpoint of CD, Ao and Bo bisect ∠ DAB, ∠ ABC, ∠ AOB = 120 ° respectively, and prove AD + Half DC + BC = ab

In the quadrilateral ABCD, point O is the midpoint of CD, Ao and Bo bisect ∠ DAB, ∠ ABC, ∠ AOB = 120 ° respectively, and prove AD + Half DC + BC = ab

Make de ⊥ Ao, intersect AB with E; make CF ⊥ Bo, intersect AB with F; connect OE, of
∵ AO and Bo are equally divided into ∠ bad and ∠ ABC
It is obvious that △ ade is isosceles triangle, that is AE = ad
Δ BCF is an isosceles triangle;: BF = BC
△OAD≡△OAE (SAS：OA=OA),
Δ OCB ≡ OCF (similarly)
So: od = OE, OC = of,
∠AOD=∠AOE,∠BOC=∠BOF
Point O is the midpoint of CD, that is, OE = of
∵ ∠AOB=120°
∴ ∠AOD+∠COB=60°
It has been proved that: ∠ AOD = ∠ AOE, ∠ BOC = ∠ BOF
∴ ∠AOE+∠COF=60°
∴ ∠FOE=60°
∵ OF=OE
The ∧ foe is an equilateral triangle
∴ FE=OF=OC=1/2DC
It has been proved that BC = BF, ad = AE
∴ AD+1/2DC+BC=AE+EF+FB=AB

Quadrilateral ABCD, ∠ D + ∠ DAB = 240, Ao bisection ∠ DAB, Bo bisection ∠ ABC, find the degree of ∠ BOC

∵∠D+∠DAB=240
Ao bisection ∠ DAB, Bo bisection ∠ ABC
∴∠BAC+∠ABO=120
∴∠BOC=120

Crazy guess picture, there is a black semicircle on the left, green arrow on the right, is a brand and English, what is it

AMD.

What figure can the upper circle and the lower semicircle be folded into?

The stack should be a cone, and the arc side length of the lower semicircle should be equal to the circumference of the upper circle

"A small circle on the top and a semicircle with radius equal to the diameter of the previous circle on the bottom" is an expanded view of what solid figure is

Cone

A three-dimensional figure, viewed from the front, has four cubes above and four cubes below. Viewed from the left, it has two cubes above and two cubes below. At least () cubes are needed
You need () cubes at most. The first figure looks like this
0000 the second one like this 00
00

A three-dimensional figure, from the front view, is four small cubes above and four small cubes below. From the left view, it is two small cubes above and two small cubes below. At least (10) small cubes are required
4+4+2 = 10

A three-dimensional figure composed of the same small cube, three views are 3 * 3 squares,
Among them, small cubes can't be suspended. How many such cubes do you need? If they can be suspended? Fast, urgent

Normal three small squares, and then three layers, and then from the upper right corner 2x2 can only need one layer, the lower left corner only needs one layer, so 27-10 = 17

The two three-dimensional figures are made up of three small cubes, and the number of squares seen from () is the same
The two three-dimensional figures are made up of three small cubes, and the number of squares seen from () is the same
A. Top B, side C, front

It's supposed to be a

Use 7 same small cubes to build a three-dimensional figure. From the front, back, left, top and right, there are 5 squares. How to build? How many do you need?

Take a cube as the center, and place one cube on each of the six sides of the cube, which meets the conditions you said. So there are seven

For a three-dimensional image, you can see five small squares on the top and two squares on the right. If you want to build such a three-dimensional image, how many small cubes do you need?

5