Is the sum of squares of the lengths of the three middle lines of a triangle really equal to 3 / 4 of the sum of squares of the lengths of its three sides? Be brief and easy to understand

Is the sum of squares of the lengths of the three middle lines of a triangle really equal to 3 / 4 of the sum of squares of the lengths of its three sides? Be brief and easy to understand

The proposition holds
As shown in the picture
prove:
AB^2=AD^2+BC^2/4-2*AD*BC/2*COS∠BDA
AC^2=AD^2+BC^2/4-2*AD*BC/2*COS(180°-∠BDA)
have to
AB^2+AC^2=2AD^2+BC^2/2   (1)
Similarly:
BC^2+AC^2=2CF^2+AB^2/2   (2)
AB^2+BC^2=2BE^2+AC^2/2   (3)
It is obtained by adding equations (1), (2) and (3)
AD^2+CF^2+BE^2=3/4(AB^2+AC^2+BC^2)

X/(X-1)

x/(x-1)

There is a flagpole on a rectangular playground of a school. The distance from the bottom of the flagpole to the three corners of the playground is 30cm, 40cm and 50cm in turn, then the distance from the bottom of the flagpole to the fourth corner
The distance between them is

Make two lines parallel to the two sides of the rectangle through the flagpole, divide the original rectangle into four small rectangles, and get the result according to the Pythagorean theorem
30 square + x square = 40 square + 50 square
32 under x = 10 radical

2,3,5,8,13,,,
emergency

21.34

How to calculate X-1 / 5x = 4

4/5x=4 x=4×5/4=5

The basketball court is 28m long and 15m wide. Draw its plan with a scale of 1:500 (only draw the boundary). Just tell me how many cm the length and width should be drawn

1: 500 means one millimeter, 0.5 meters
After conversion, the length is 5.6CM and the width is 3cm

In the isosceles right triangle ABC, ∠ a = 90 °, BC = 6cm, BD bisection ∠ ABC intersects AC at D, de ⊥ BC at e, then the perimeter of △ CDE is______ ．

As shown in the figure, ∵ ABC is isosceles right triangle, ∵ AC = AB, let AB = x, then 2x2 = BC2 = 36, the solution is x = 32, ∵ BD bisection, ∵ ABC intersects AC at D, de ⊥ BC at e, ∵ ad = De, in RT △ abd and RT △ EBD,