3. In right-angled trapezoidal ABCD, if AD//BC, angle ABC=90 degrees, AD:BC= 1:4, BD: AC = _ _ _ _ _ _.
13. It is known that the upper base length of a trapezoid is 2, the lower base length is 5, and one waist length is 4, so the range of the other waist length is _ _.
It is known that A, B and C are three sides of a triangle, and the square of (a+b+c) is equal to 3 (A +b +c). Prove that this triangle is an equilateral triangle.
Answer: 60
If you do two heights, you can get a rectangle. The bottom is divided into three parts: 1, 1, 1. A right triangle with the waist as the hypotenuse has a cotangent of 2, so the base angle is 60.
In the right-angled trapezoidal ABCD, AD//BC, angle ABC=90 degrees, AD:BC= 1:4, then BD:AC= 1:2.
Solution: in trapezoidal ABCD, AD‖BC, and ∠ A = 90, AD = 1, BC = 4,
Derived from Pythagorean theorem
AB^2=BD^2-AD^2=AC^2-BC^2
That is, AC 2-BD 2 = BC 2-AD 2 =16-1=15 (1).
If the extension line of DE‖AC intersects BC at E, ACED is a parallelogram, CE=AD= 1, DE=AC,
BE=BC+CE=4+ 1=5
AC⊥BD,∠BDE=90
From Pythagorean Theorem
DE^2+BD^2=BE^2
That is, AC 2+BD 2 = 5 2 = 25 (2)
For (1)+(2), 2ac 2 = 40.
AC^2=20,BD^2=5
AC^2:BD^2=20:5=4: 1
AC:BD=2: 1
13. It is known that the upper base length of a trapezoid is 2, the lower base length is 5, and one waist length is 4, so the range of the other waist length is _ _.
1 & lt； M< seven
When the waist is known to translate to the other end of the upper sole, the part of the lower sole that is longer than the lower sole is 5-2 = 3, which forms a triangle with this waist 4 and waist M, so: 4-3.
Namely: 1
(a+b+c)^2
=a^2+b^2+c^2+2ab+2bc+2ca
=3(a^2+b^2+c^2)
The arrangement (a-b) 2+(b-c) 2+(c-a) 2 = 0.
So a=b=c, so it is an equilateral triangle.