Question

В треугольнике АВС точка М принадлежит стороне ВС. Отрезки AN и СМ пересекаются в точке Р. Прямая ВР пересекает сторону АС в точке R. Какую часть АС составляет отрезок AR, если АМ : МВ=3 : 5 и NC : BN = 3 : 2?

Answer 1

Ответ: AR=2/7*AC
Решение в файле.
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Answer 2

Triangle ABC and Point M

In triangle ABC, point M lies on side BC. The segments AN and CM intersect at point P. The line VR intersects side AC at point R. We are asked to determine the portion of segment AC that is formed by segment AR, given that AM:MB = 3:5 and NC:NB = 3:2.

To solve this problem, we can use the concept of similar triangles and proportions.

Proportions and Similar Triangles

Let's consider the triangles AMR and BNR. These triangles are similar because they share an angle at point R and have proportional sides.

From the given information, we know that AM:MB = 3:5 and NC:NB = 3:2. This means that the lengths of AM and NC are 3x and 3y, respectively, and the lengths of MB and NB are 5x and 2y, respectively.

Using the concept of similar triangles, we can set up the following proportion:

AR:RB = AM:MB

Substituting the lengths we found earlier, we have:

AR:RB = 3x:5x

Simplifying the proportion, we get:

AR:RB = 3:5

This means that the length of segment AR is 3/8 of the length of segment AC, and the length of segment RB is 5/8 of the length of segment AC.

Therefore, segment AR constitutes 3/8 of segment AC.

Conclusion

In triangle ABC, with point M on side BC, the portion of segment AC formed by segment AR is 3/8 of segment AC.