
2). В треугольнике СDE точка М лежит на стороне СЕ, причём - острый. Докажите, что DE > DM. И
нарисуйте пожалуйста треугольник к задаче. Заранее спасибо.

Ответы на вопрос




Problem Statement
We are given a triangle CDE, where point M lies on the side CE and is acute. We need to prove that DE > DM.Proof
To prove that DE > DM, we can use the triangle inequality theorem. According to this theorem, the sum of the lengths of any two sides of a triangle must be greater than the length of the third side.In our case, we have triangle CDE, where DE is one side and DM is another side. We need to show that the sum of the lengths of DE and DM is greater than the length of the remaining side, which is CE.
Let's assume that DE ≤ DM. In this case, we can write the following inequality:
DE + DM ≤ DM + DM
Simplifying the inequality, we get:
DE + DM ≤ 2DM
Now, let's consider triangle CDM. 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 CDM, we have:
DM + CM > CD
Since CM is a part of CE, we can substitute CM with CE - ME:
DM + CE - ME > CD
Rearranging the inequality, we get:
CE - ME > CD - DM
Since CE = CD + DE, we can substitute CE with CD + DE:
CD + DE - ME > CD - DM
Simplifying the inequality, we have:
DE - ME > -DM
Adding ME to both sides of the inequality, we get:
DE > ME - DM
But we know that ME > 0, since M lies on the side CE. Therefore, we can conclude that:
DE > ME - DM
Since ME - DM is a positive value, we can further simplify the inequality to:
DE > DM
Thus, we have proved that DE > DM.
Triangle Diagram
Here is a diagram of triangle CDE:``` C / \ / \ / \ / \ D---------E | M ```
In the diagram, C, D, and E represent the vertices of the triangle, and M represents the point on side CE.
Please note that the diagram is not to scale and is only for illustrative purposes.
I hope this explanation and diagram help you understand the proof. Let me know if you have any further questions!


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