Question

В параллелограмме АВСД точка Е-Середина стороны СД. Известно,что ЕА=ЕВ.Докажите,что данный параллелограмм -прямоугольник. Помогите пожалуйста нужно полное решение:*

Answer 1

Пусть BC=AD=b; AB=CD=a; CE=ED=a/2/. По теореме косинусов имеем:
BE^2=b^2+a^2/4-2*b*a/2*cosC=b^2+a^2/4-b*a*cosC;

AD^2=b^2+a^2/4-2*b*a/2*cosD=b^2+a^2/4-b*a*cosD;
По условию BE=AD. Значит, BE^2=AD^2, т.е.

b^2+a^2/4-b*a*cosC=b^2+a^2/4-b*a*cosD
cosC=cosD, значит C=D и C+D=180 как сумма углов, прилежащих к одной стороне паралл.
Отсюда следует, что C=D=90

Answer 2

Параллелограмм АВСД, СЕ=ЕД, АЕ=ВЕ

Проводим ЕН - линия параллельна АД делит АВ на две равные части, АН=ВН=СД=СЕ

треугольник АЕВ равнобедренный, ЕН - медиана, высота , биссектириса

ЕН перпендикулярен АВ . Значит АД и ВС также перпендикулярены

Теорема - если в параллелограмме есть прямой угол, то фигура прямоугольник

Answer 3

Given Information:

In the parallelogram ABCD, point E is the midpoint of side CD, and it is known that EA = EB.

Proof:

To prove that the given parallelogram is a rectangle, we need to show that one of its angles is a right angle.

Let's consider triangle AED. Since E is the midpoint of CD, we can conclude that DE = EC Additionally, we know that EA = EB. Therefore, triangle AED is an isosceles triangle with two equal sides, AE and DE.

By the Converse of the Isosceles Triangle Theorem, if a triangle has two sides that are congruent, then the angles opposite those sides are congruent. In this case, angle AED is congruent to angle ADE.

Now, let's consider triangle ABE. We know that EA = EB, and angle AEB is congruent to angle AED (opposite angles of a parallelogram are congruent). Therefore, triangle ABE is also an isosceles triangle with two equal sides, AE and BE.

By the Converse of the Isosceles Triangle Theorem, angle ABE is congruent to angle BAE.

Since angle AED is congruent to angle ABE, and angle ABE is congruent to angle BAE, we can conclude that angle AED is congruent to angle BAE.

Now, let's consider the sum of angles AED and BAE. Since they are congruent, their sum is 180 degrees.

Therefore, angle AED + angle BAE = 180 degrees.

Since angle AED is congruent to angle BAE, we can rewrite the equation as:

2 * angle AED = 180 degrees.

Dividing both sides of the equation by 2, we get:

angle AED = 90 degrees.

Thus, we have proved that angle AED is a right angle, which means that the given parallelogram ABCD is a rectangle.

Conclusion:

The given parallelogram ABCD is a rectangle, as proved above.