Prove: the theorem of the middle line of triangle (that is, the sum of two waist squares is twice, equal to the sum of the square of the bottom edge and the square of the middle line of that edge is four times) yes

Prove: the theorem of the middle line of triangle (that is, the sum of two waist squares is twice, equal to the sum of the square of the bottom edge and the square of the middle line of that edge is four times) yes

Let △ ABC, ad be the middle line on the edge of BC, BD = CD = BC / 2, and let AB & sup2; = ad & sup2; + BD & sup2; - 2ad * BD * cos (1)AC²=AD²+DC²-2AD*CD*cos…… (2)∵BD=CD=BC/2,∴(1)+(2):AB²+AC²=2AD²+2(BC/2)²...

Can we prove that a triangle is a right triangle with HL theorem

No, using HL theorem is a special theory to prove the congruence of two right triangles

It weighs 110 grams per square meter, the door width is 1.5 meters, the quantity is 900 meters, and the unit price is 28 yuan / kg. How can I convert it into rice price?

462 yuan / M

As shown in the figure, if there are two adjacent squares in the rectangle, and the areas are 2 and 4 respectively, the area of the shadow part is_____ -

The side length of large square is 4 = 2 × 2
Side length of small square 2 = √ 2 ×√ 2
So the shadow is (2 - √ 2) ×√ 2 = 2 √ 2-2

What are the axisymmetric figures in life?
Like a table

Generally, as long as it is a cylinder, cone, ball, cube, cuboid, geometry are axisymmetric graphics, so, in life, there are cups (no handle), books, volleyball, football, basketball, badminton racket, lamp, cabinet, fan, stool, table, bed, quilt, sofa, couplet, pencil box, etc
There are airplanes, butterflies, pine trees, Bank of China logo, industrial and Commercial Bank logo, five stars on the red flag

Given that 2007 (x + y) square and 2008 / 1 / 2y-0.5 / are opposite numbers, find the value of X and y

x=-1 y=1

Let a and B satisfy a + B = A-B and prove a ⊥ B

Xiaolian, let me help you. Hee hee, although I don't learn very well
Because | a + B | = | a-b|
So (a + b) * (a + b) = (a-b) * (a-b)
A & sup2; + 2A · B + B & sup2; = A & sup2; - 2A · B + B & sup2;
It is reduced to 4A · B = 0
So a · B = 0
That is a ⊥ B

As shown in the figure, in △ ABC, D and E are the midpoint of BC and ad respectively, and the area of △ Abe is 1, so find the area of △ ABC
Be quick

 
There are pictures in the document: the area of △ Abe is 1, e is the midpoint of AD, then the area of △ abd is 2;
Similarly, the area of △ abd is 2 & nbsp;, & nbsp;, then the area of △ ABC is 2 & nbsp; x2 = 4 & nbsp;
 

The minimum value of the function y = X3 / 3 + x2-3x-4 on [0,2] is

y'=x²+2x-3=0
x=-3,x=1
Then 0

High one mathematics, vector collinear problem proof
Let the vector OA = E1, OB = E2, OC = E3, if there are real numbers λ 1, λ 2, λ 3 which are not all zero, such that λ 1E1 + λ 2e2 + λ 3e3 = 0, and λ 1 + λ 2 + λ 3 = 0, try to prove that ABC three points are collinear
Trouble process to write in detail, online, etc., thank you

λ1e1+λ2e2+λ3e3=0
That is, λ 1oa + λ 2ob + (- λ 1 - λ 2) OC = 0
So λ 1 (oa-oc) + λ 2 (ob-oc) = 0
That is, λ 1ca = - λ 2CB
So Ca and CB are collinear, that is, a, B and C are collinear