Question

На стороне ac треугольника abc взяли точку m так,что am:cm=2:3. На отрезке BM взяли точку K так,чтобы BK:MK=3:4. В каком отношении прямая AK делит сторону BC треугольника? Надо использовать теорему Фалеса/пропорциональных отрезков

Answer 1

Ответ:

BE : EC = 3 : 10

Объяснение:

АМ : МС = 2 : 3

ВК : КМ = 3 : 4

Проведем МР║АЕ.

Для угла МВС по обобщенной теореме Фалеса:

$\dfrac{BE}{EP}=\dfrac{BK}{KM}=\dfrac{3}{4}$

Для угла ЕАС по обобщенной теореме Фалеса:

$\dfrac{PC}{EP}=\dfrac{MC}{AM}=\dfrac{3}{2}=\dfrac{6}{4}$

То есть, ЕР составляет 4 части, ВЕ составляет 3 части, а РС - 6 частей.

EC = EP + PC = 4 + 6 = 10 частей

$\dfrac{BE}{EC}=\dfrac{3}{10}$

Answer 2

Problem Analysis

We are given triangle ABC, and a point M on side AC such that AM:CM = 2:3. We are also given that point K is on segment BM such that BK:MK = 3:4. We need to determine the ratio in which line AK divides side BC of the triangle.

Solution

To solve this problem, we can use the theorem of Thales or the property of proportional segments.

According to the theorem of Thales, if a line is drawn parallel to one side of a triangle, it divides the other two sides proportionally. In this case, we can draw a line parallel to side BC through point A, and it will intersect side BM at point K.

Let's assume that line AK divides side BC in the ratio x:y. This means that the length of segment BK is x times the length of segment KC. We can set up the following proportion:

BK:KC = x:y

We are given that BK:MK = 3:4. Since line AK is parallel to side BC, we can use the property of proportional segments to establish the following relationship:

BK:KC = AM:MC

Substituting the given values, we have:

3:4 = 2:3

To solve this proportion, we can cross-multiply:

3 * 3 = 4 * 2

9 = 8

This is not a true statement, which means that our assumption that line AK divides side BC in the ratio x:y is incorrect.

Therefore, we can conclude that line AK does not divide side BC of triangle ABC in any specific ratio.

Answer

The line AK does not divide side BC of triangle ABC in any specific ratio.