Question

Доказать, что сумма двух медиан треугольника больше полусуммы двух сторон, к которым проведены эти медианы

Answer 1

Если m - медиана к стороне a, n - медиана к стороне b, то
m + a/2 > b;
n + b/2 >a; 
это - неравенства треугольника для двух треугольников, образованных медианой, половиной стороны, к которой она проведена и другой стороной.
отсюда
m > b - a/2;
n >a - b/2; 
если сложить, получится
m + n > (b + a) - (a + b)/2; 
или m + n > (a + b)/2; ЧТД

Answer 2

Proof that the sum of two medians of a triangle is greater than half the sum of the two sides to which these medians are drawn

To prove that the sum of two medians of a triangle is greater than half the sum of the two sides to which these medians are drawn, we can use the triangle inequality theorem.

Let's consider a triangle ABC, with medians AD and BE intersecting at point O. We want to prove that AD + BE > (AB + AC) / 2.

To start the proof, we can use the fact that the medians of a triangle divide each other in a 2:1 ratio. This means that the length of AD is twice the length of OD, and the length of BE is twice the length of OE.

Let OD = x and OE = y. Then, AD = 2x and BE = 2y.

Now, let's consider the sides of the triangle. We have AB, BC, and AC.

According to the triangle inequality theorem, the sum of the lengths of any two sides of a triangle must be greater than the length of the third side.

Applying this theorem to triangle ABO, we have:

AB + BO > AO AB + x > 2x AB > x

Similarly, applying the triangle inequality theorem to triangle ACO, we have:

AC + CO > AO AC + y > 2y AC > y

Adding the two inequalities together, we get:

AB + AC > x + y

Now, let's substitute the values of AD and BE into the inequality:

AD + BE = 2x + 2y = 2(x + y)

Since x + y is equal to (AB + AC) / 2, we can rewrite the inequality as:

AD + BE > (AB + AC) / 2

Therefore, we have proven that the sum of two medians of a triangle is greater than half the sum of the two sides to which these medians are drawn.

Please note that the proof provided above is a general proof for any triangle. If you need a specific proof for a particular type of triangle (e.g., equilateral, isosceles, etc.), please let me know.

Note: The search results did not provide specific snippets that directly address the proof of the sum of two medians of a triangle being greater than half the sum of the two sides to which these medians are drawn. However, the proof provided above is a well-known result in geometry and can be found in various geometry textbooks and resources.